NCERT Solutions for Class 12 Physics Chapter 11 - Dual Nature of Radiation and Matter

NCERT Solutions for Class 12 Physics Chapter 11 - Dual Nature of Radiation and Matter

Edited By Vishal kumar | Updated on Apr 30, 2025 02:02 PM IST | #CBSE Class 12th

Ever thought how light can be both a wave and a particle? That is the magic of dual nature of radiation and matter and it is one of the most interesting topics in physics, On this page, you will find simple and clear Class 12 Physics Chapter 11 NCERT solutions,created by experts faculty and perfect for quick revision. You can even download the PDF to study offline, anytime, anywhere.

NCERT solution for Class 12 Chapter 11 Dual Nature of Radiation and Matter, you will explore both the wave and particle nature of light and matter. One of the most important topics covered is the photoelectric effect, which has been asked about frequently in CBSE board exams. This chapter explains how light sometimes behaves like a wave and sometimes like a particle, depending on the experiment.

Try solving all NCERT questions on your own first. If you get stuck, refer to the NCERT solution for Class 12 Physics Chapter 11 solutions for clear explanations. Also, make sure to understand all the graphs in this chapter, as they are important for solving numerical problems and developing a strong interest in the topic.

This Story also Contains
  1. NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter: Additional Questions
  2. Class 12 physics NCERT Chapter 11: Higher Order Thinking Skills (HOTS) Questions
  3. Approach to Solve Questions of Dual Nature of Matter and Radiation Class 12
  4. What Extra Should Students Study Beyond NCERT for JEE/NEET?
  5. Class 12 Physics Chapter 11 NCERT Solutions: Important Formulas
  6. Dual Nature of Matter and Radiation Class 12-Topics
  7. NCERT Solutions for Class 12 Physics: Chapter-Wise
  8. Also, check NCERT Books and NCERT Syllabus here:
NCERT Solutions for Class 12 Physics Chapter 11 - Dual Nature of Radiation and Matter
NCERT Solutions for Class 12 Physics Chapter 11 - Dual Nature of Radiation and Matter
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Download NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter clear, step-by-step answers to help you understand concepts like photoelectric effect and particle-wave duality with ease. Perfect for quick revision and JEE/NEET preparation.

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Class 12 Physics Dual Nature of Radiation and Matter - Exercise Solutions

11.1 (a) Find the maximum frequency of X-rays produced by $\small 30 \hspace{1mm}kV$ electrons.

Answer:

The X-Rays produced by electrons of 30 keV will have a maximum energy of 30 keV.

By relation,

$\begin{aligned} & e V_0=h \nu \\ & \nu=\frac{e V_0}{h} \\ & \nu=\frac{1.6 \times 10^{-19} \times 30 \times 10^3}{6.62 \times 10^{-34}} \\ & \nu=7.25 \times 10^{18} \mathrm{~Hz}\end{aligned}$

11.1 (b) Find the minimum wavelength of X-rays produced by $\small 30 \hspace{1mm}kV$ electrons.

Answer:

From the relation $eV_{0}=h\nu$, we have calculated the value of frequency in the previous questions, using that value and the following relation

$\lambda =\frac{c}{\nu }$
$ \lambda =\frac{3\times 10^{8}}{7.25\times 10^{18}}$
$ \lambda =0.04\ nm$

11.2 (a) The work function of caesium metal is $\small 2.14\hspace{1mm}eV$. When light of frequency $6 \times 10^{14} \mathrm{~Hz}$ is incident on the metal surface, photoemission of electrons occurs. What is the maximum kinetic energy of the emitted electrons?

Answer:

The energy of the incident photons is E is given by

$E=h\nu $
$E=\frac{6.62\times 10^{-34}\times 6\times 10^{14}}{1.6\times 10^{-19}}$
$ E=2.48\ eV$

Maximum Kinetic Energy is given by

$KE_{max}=E-\phi _{0}$
$ KE_{max}=2.48-2.14$
$ KE_{max}=0.34\ eV$

11.2 (b) The work function of caesium metal is $2.14\hspace{1mm}eV$ When light of frequency $6 \times 10^{14} \mathrm{~Hz}$ is incident on the metal surface, photoemission of electrons occurs. What is the stopping potential

Answer:

The stopping potential depends on the maximum Kinetic Energy of the emitted electrons. Since maximum Kinetic energy is equal to 0.34 eV, stopping potential is the maximum kinetic energy by charge equal to 0.34 V.

11.2 (c) The work function of caesium metal is $2.14\hspace{1mm}eV$ . When light of frequency $6 \times 10^{14} \mathrm{~Hz}$ is incident on the metal surface, photoemission of electrons occurs. What is the maximum speed of the emitted photoelectrons?

Answer:

The electrons with the maximum kinetic energy of 0.34 eV will have the maximum speed

$\begin{aligned} & K E_{\max }=0.34 \mathrm{eV} \\ & K E_{\max }=5.44 \times 10^{-20} \mathrm{~J} \\ & v_{\max }=\sqrt{\frac{2 K E_{\max }}{m}} \\ & v_{\max }=\sqrt{\frac{2 \times 5.44 \times 10^{-20}}{9.1 \times 10^{-31}}} \\ & v_{\max }=3.44 \times 10^5 \mathrm{~ms}^{-1}\end{aligned}$

11.3 The photoelectric cut-off voltage in a certain experiment is $1.5\hspace{1mm}V$. What is the maximum kinetic energy of photoelectrons emitted?

Answer:

Since the photoelectric cut-off voltage is 1.5 V. The maximum Kinetic Energy (eV) of photoelectrons emitted would be 1.5 eV.

KE max =1.5 eV

KE max =2.4 $\times$ 10 -19 J

11.4 (a) Monochromatic light of wavelength $632.8\hspace{1mm} nm$ is produced by a helium-neon laser. The power emitted is $9.42\hspace{1mm} mW$ . Find the energy and momentum of each photon in the light beam.

Answer:

The energy of photons is given by the relation

$\begin{aligned}
& E=h \nu \\
& E=\frac{h c}{\lambda} \\
& E=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{632 \times 10^{-9}} \\
& E=3.14 \times 10^{-19} J
\end{aligned}$

Momentum is given by De Broglie's Equation

$\begin{aligned}
p & =\frac{h}{\lambda} \\
p & =\frac{6.62 \times 10^{-34}}{632.8 \times 10^{-9}} \\
p & =1.046 \times 10^{-27} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}$

The energy of the photons in the light beam is $3.14 \times 10^{-19} \mathrm{~J}$ and the momentum of the photons is $1.046 \times 10^{-27} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$.

11.4 (b) Monochromatic light of wavelength $632.8\hspace{1mm}nm$ is produced by a helium-neon laser. The power emitted is $9.42\hspace{1mm}mW$ .

How many photons per second, on the average, arrive at a target irradiated by this beam? Assume the beam to have uniform cross-section which is less than the target area),

Answer:

Power of the light beam, P =9.42 mW

If n number of photons arrive at a target per second nE=P (E is the energy of one photon)

$n=\frac{P}{E}$
$n=\frac{9.42\times 10^{-3}}{3.14\times 10^{-19}}$
$ n=3\times 10^{16}$

11.4 (c) Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW. How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?

Answer:

Mass of Hydrogen Atom (m)=1.67 $\times$ 10 -27 kg.

The speed at which hydrogen atom must travel to have momentum equal to that of the photons in the beam is v given by

$v=\frac{p}{m}$
$v=\frac{1.05\times 10^{-27}}{1.67\times 10^{-27}}$
$ v=0.628\ ms^{-1}$

11.5 In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 * 10-15 Vs . Calculate the value of Planck’s constant.

Answer:

The slope of the cut-off voltage versus frequency of incident light is given by h/e where h is Plank's constant and e is an electronic charge.

$h=slope\times e$

$h=4.12\times10^{-15}\times1.6\times10^{-19}$

$h=6.59210^{-34} Js$

11.6 The threshold frequency for a certain metal is $3.3 \times 10^{14} \mathrm{~Hz}$ . If light of frequency$8.2 \times 10^{14} \mathrm{~Hz}$ is incident on the metal, predict the cutoff voltage for the photoelectric emission.

Answer:

Threshold frequency of the given metal $\left(\nu_0\right)=3.3 \times 10^{14} \mathrm{~Hz}$
The work function of the given metal is

$\begin{aligned}
\phi_0 & =h \nu_0 \\
\phi_0 & =6.62 \times 10^{-34} \times 3.3 \times 10^{-14} \\
\phi_0 & =2.18 \times 10^{-19} \mathrm{~J}
\end{aligned}$

The energy of the incident photons

$\begin{aligned}
& E=h \nu \\
& E=6.62 \times 10^{-34} \times 8.2 \times 10^{14} \\
& E=5.42 \times 10^{-19} \mathrm{~J}
\end{aligned}$

Maximum Kinetic Energy of the ejected photoelectrons is

$\begin{aligned}
& E-\phi_0=3.24 \times 10^{-19} J \\
& E-\phi_0=2.025 \mathrm{eV}
\end{aligned}$

Therefore the cut off voltage is 2.025 eV

11.7 The work function for a certain metal is $4.2\hspace{2mm}eV$ . Will this metal give photoelectric emission for incident radiation of wavelength $330\hspace{1mm}nm$ ?

Answer:

The energy of photons having 330 nm is

$\\E=\frac{hc}{\lambda }$
$ E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{330\times 10^{-9}\times 1.6\times 10^{-19}}$
$ E=3.7\ eV$

Since this is less than the work function of the metal there will be no photoelectric emission.

11.8 Light of frequency $7.21 \times 10^{14} \mathrm{~Hz}$ is incident on a metal surface. Electrons with a maximum speed of $6.0\times 10^5\hspace{1mm}m/s$ are ejected from the surface. What is the threshold frequency for photoemission of electrons?

Answer:

The energy of incident photons is E given by
$\begin{aligned}
& E=h \nu \\
& E=6.62 \times 10^{-34} \times 7.21 \times 10^{14} \\
& E=4.77 \times 10^{-19} J
\end{aligned}$

Maximum Kinetic Energy of ejected electrons is
$\begin{aligned}
& K E_{\max }=\frac{1}{2} m v^2 \\
& K E_{\max }=\frac{9.1 \times 10^{-31} \times\left(6 \times 10^5\right)^2}{22^2} \\
& K E_{\max }=1.64 \times 10^{-19} \mathrm{~J}
\end{aligned}$

Work Function of the given metal is
$\phi_0=E-K E_{\max }=3.13 \times 10^{-19} \mathrm{~J}$

The threshold frequency is therefore given by
$\begin{aligned}
& \nu_0=\frac{\phi_0}{h} \\
& \nu_0=4.728 \times 10^{14} \mathrm{~Hz}
\end{aligned}$

11.9 Light of wavelength $488\hspace{1mm}nm$ is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is $0.38\hspace{1mm}V$ . Find the work function of the material from which the emitter is made.

Answer:

The energy of incident photons is given by

$E=\frac{hc}{\lambda }$
$E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{488\times 10^{-9}\times 1.6\times 10^{-19}}$
$ E=2.54\ eV$

Cut-off potential is 0.38 eV

Therefore, work function is, 2.54-0.38= 2.16 eV

11.10 (a) What is the de Broglie wavelength of a bullet of mass $0.040\hspace{1mm}kg$ travelling at the speed of $1.0\hspace{1mm}km/s$.

Answer:

The momentum of the bullet is
$\begin{aligned}
& p=m v \\
& p=0.04 \times 10^3 \\
& p=40 \mathrm{~kg~m} \mathrm{~s}^{-1}
\end{aligned}$

De Broglie wavelength is
$\begin{aligned}
\lambda & =\frac{h}{p} \\
\lambda & =\frac{6.62 \times 10^{-34}}{40} \\
\lambda & =1.655 \times 10^{-35} \mathrm{~m}
\end{aligned}$

(b) What is the de Broglie wavelength of a ball of mass $0.060\hspace{1mm}kg$ moving at a speed of $1.0\hspace{1mm}m/s$ .

Answer:

The momentum of the ball is

$\begin{aligned}
& p=m v \\
& p=0.06 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}$

De Broglie wavelength is

$\begin{aligned}
\lambda & =\frac{h}{p} \\
\lambda & =\frac{6.62 \times 10^{-34}}{0.06} \\
\lambda & =1.1 \times 10^{-32} \mathrm{~m}
\end{aligned}$

(c) What is the de Broglie wavelength of a dust particle of mass $1.0 \times 10^{-9} \mathrm{~kg}$ drifting with a speed of $2.2\hspace{1mm}m/s$?

Answer:

The momentum of the dust particle is
$\begin{aligned}
& p=m v \\
& p=10^{-9} \times 2.2 \\
& p=2.2 \times 10^{-9} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}$

De Broglie wavelength is
$\begin{aligned}
\lambda & =\frac{h}{p} \\
\lambda & =\frac{6.62 \times 10^{-34}}{2.2 \times 10^{-9}} \\
\lambda & =3.01 \times 10^{-25} \mathrm{~m}
\end{aligned}$

11.11 Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

Answer:

For a photon we know that it's momentum $(\mathrm{p})$ and Energy $(\mathrm{E})$ are related by following equation
$\mathrm{E}=\mathrm{pc}$

We also know
$E=h \nu$

Therefore the De Broglie wavelength is
$\begin{aligned}
\lambda & =\frac{h}{p} \\
\lambda & =\frac{h}{E / c} \\
\lambda & =\frac{h c}{h \nu} \\
\lambda & =\frac{c}{\nu}
\end{aligned}$

The above de Broglie wavelength is equal to the wavelength of electromagnetic radiation.

NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter: Additional Questions

1.a) An electron and a photon each have a wavelength of $1.00\hspace{1mm}nm$ . Find their momenta.

Answer:

Their momenta depend only on the de Broglie wavelength, therefore, it will be the same for both the electron and the photon

$\begin{aligned}
p & =\frac{h}{\lambda} \\
p & =\frac{6.62 \times 10^{-34}}{10^{-9}} \\
p & =6.62 \times 10^{-25} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}$

b) An electron and a photon each have a wavelength of $1.00\hspace{1mm}nm$ . Find the energy of the photon.

Answer:

The energy of the photon is given by

$\\E=\frac{hc}{\lambda }\\$

h is the Planks constant, c is the speed of the light and lambda is the wavelength

$E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{10^{-9}}\\$

$E=1.86\times 10^{-16}\ J$

c) An electron and a photon each have a wavelength of $1.00\hspace{1mm}nm$ . Find the kinetic energy of electron.

Answer:

The kinetic energy of the electron is. In the below equation p is the momentum

$K=\frac{p^{2}}{2m_{e}}$
$K=\frac{(6.62\times 10^{-25})^{2}}{2\times 9.1\times 10^{-31}}$
$K=2.41\times 10^{-19}\ J$

2.a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.4*10-10 m?

Answer:

For the given wavelength momentum of the neutron will be $p$ given by
$\begin{aligned}
p & =\frac{h}{\lambda} \\
p & =\frac{6.62 \times 10^{-34}}{1.4 \times 10^{-10}} \\
p & =4.728 \times 10^{-24} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}$

The kinetic energy K would therefore be
$\begin{aligned}
K & =\frac{p^2}{2 m} \\
K & =\frac{\left(4.728 \times 10^{-24}\right)^2}{2 \times 1.675 \times 10^{-27}} \\
K & =6.67 \times 10^{-21} J
\end{aligned}$

b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of $(3/2)\hspace{1mm}kT$ at $300\hspace{1mm}K$ .

Answer:

The kinetic energy of the neutron is
$\begin{aligned}
& K=\frac{3}{2} k T \\
& K=\frac{3}{2} \times 1.38 \times 10^{-23} \times 300 \\
& K=6.21 \times 10^{-21} J
\end{aligned}$

Where k Boltzmann's Constant is $1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$
The momentum of the neutron will be $p$
$\begin{aligned}
& p=\sqrt{2 m_N K} \\
& p=\sqrt{2 \times 1.675 \times 10^{-27} \times 6.21 \times 10^{-21}} \\
& p=4.56 \times 10^{-24} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}$

Associated De Broglie wavelength is
$\begin{aligned}
\lambda & =\frac{h}{p} \\
\lambda & =\frac{6.62 \times 10^{-34}}{4.56 \times 10^{-24}} \\
\lambda & =1.45 \times 10^{-10} \mathrm{~m}
\end{aligned}$

De Broglie wavelength of the neutron is 0.145 nm .

4. What is the de Broglie wavelength of a nitrogen molecule in air at $300\hspace{1mm}K$ ? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen $=14.0076\hspace{1mm} \mu$ )

Answer:

Since the molecule is moving with the root-mean-square speed the kinetic energy $K$ will be given by $\mathrm{K}=3 / 2 \mathrm{kT}$ where k is the Boltzmann's constant and T is the absolute Temperature

In the given case Kinetic Energy of a Nitrogen molecule will be
$\begin{aligned}
& K=\frac{3}{2} \times 1.38 \times 10^{-23} \times 300 \\
& K=6.21 \times 10^{-21} J
\end{aligned}$

$\text { Mass of Nitrogen molecule }=2 \times 14.0076 \times 1.66 \times 10^{-27}=4.65 \times 10^{-26} \mathrm{~kg}$

The momentum of the molecule is
$\begin{aligned}
& p=\sqrt{2 m K} \\
& p=\sqrt{2 \times 4.65 \times 10^{-26} \times 6.21 \times 10^{-21}} \\
& p=2.4 \times 10^{-23} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}$

Associated De Broglie wavelength is
$\begin{aligned}
\lambda & =\frac{h}{p} \\
\lambda & =\frac{6.62 \times 10^{-34}}{2.4 \times 10^{-23}} \\
\lambda & =2.75 \times 10^{-11} \mathrm{~m}
\end{aligned}$

The nitrogen molecule will have a De Broglie wavelength of 0.0275 nm.

5.a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of $500\hspace{1mm}V$ with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its $e/m$ is given to be $1.76 \times 10^{11} \mathrm{~kg}^{-1}$

Answer:

The kinetic energy of an electron accelerated through Potential Difference V is K=eV where e the electronic charge.

Speed of the electrons after being accelerated through a potential difference of 500 V will be

$v=\sqrt{\frac{2K}{m_{e}}}$
$v=\sqrt{\frac{2eV}{m_{e}}}$
$v=\sqrt{2\times 1.76\times 10^{11}\times 500}$
$ v=1.366\times 10^{7}ms^{-1}$
Specific charge is e/me =1.366 $\times$ 1011 C/kg

5.b) Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?

Answer:

Using the same formula we get the speed of electrons to be 1.88 $\times$ 109 m/s. This is wrong because the speed of the electron is coming out to be more than the speed of light. This discrepancy is occurring because the electron will be travelling at very large speed and in such cases(relativistic) the mass of the object cannot be taken to be the same as the rest mass.

In such a case

$m=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

where m is the relativistic mass, m 0 is the rest mass of the body, v is the very high speed at which the body is traveling and c is the speed of light.

6.a) A monoenergetic electron beam with an electron speed of $5.20 \times 10^6 \mathrm{~ms}^{-1}$ is subject to a magnetic field of $1.30 \times 10^{-4} T$ normal to the beam velocity. What is the radius of the circle traced by the beam, given $e/m$ for electron equals $1.76 \times 10^{11} \mathrm{Ckg}^{-1}$

Answer:

The force due to the magnetic field on the electron will be $F_b=e v B$ (since the angle between the velocity and magnetic field is $90^{\circ}$ )

This $\mathrm{F}_{\mathrm{b}}$ acts as the centripetal force required for circular motion. Therefore
$\begin{aligned}
& F_b=\frac{m v^2}{r} \\
& e v B=\frac{m v^2}{r} \\
& r=\frac{m v}{e B} \\
& r=\frac{5.2 \times 10^6}{1.76 \times 10^{11} \times 1.3 \times 10^{-4}} \\
& r=0.227 m
\end{aligned}$

6.b) Is the formula you employ in (a) valid for calculating radius of the path of a $20\hspace{1mm}MeV$ electron beam? If not, in what way is it modified?

Answer:

The formula used in (a) can not be used. As the electron would be travelling at a very high speed we can not take its mass to be equal to its rest mass as its motion won't be within the non-relativistic limits.

The value for the mass of the electron would get modified to

$m=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

where m is the relativistic mass, m 0 is the rest mass of the body, v is the very high speed at which the body is travelling and c is the speed of light.

The radius of the circular path would be

$r=\frac{m_{e}v}{eB\sqrt{1-\frac{v^{2}}{c^{2}}}}$

7. An electron gun with its collector at a potential of $100\hspace{1mm}V$ fires out electrons in a spherical bulb containing hydrogen gas at low pressure ( $10^{-2} \mathrm{~mm}$ of Hg). A magnetic field of $2.83\times 10^4\hspace{1mm}T$ curves the path of the electrons in a circular orbit of radius $12.0\hspace{1mm}cm$ (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method.) . Determine $e/m$ from the data.

Answer:

The kinetic energy of an electron after being accelerated through a potential difference of V volts is eV where e is the electronic charge.

The speed of the electron will become

$v=\sqrt{\frac{2eV}{m_{e}}}$

Since the magnetic field curves, the path of the electron in circular orbit the electron's velocity must be perpendicular to the magnetic field.

The force due to the magnetic field is therefore F b =evB

This magnetic force acts as a centripetal force. Therefore

$\begin{aligned}
& \frac{m_e v^2}{r}=e v B \\
& \frac{m_e v}{r}=e B \\
& \frac{m_e}{r} \times \sqrt{\frac{2 e V}{m_e}}=e B \\
& \sqrt{\frac{e}{m_e}}=\frac{\sqrt{2 V}}{B r} \\
& \frac{e}{m_e}=\frac{2 V}{r^2 B^2} \\
& \frac{e}{m_e}=\frac{2 \times 100}{\left(2.83 \times 10^{-4}\right)^2 \times(0.12)^2} \\
& \frac{e}{m_e}=1.73 \times 10^{11} C~kg^{-1}
\end{aligned}$

8.a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at $0.45\dot{A}$ . What is the maximum energy of a photon in the radiation?

Answer:

The wavelength of photons with maximum energy=0.45 $A^{\circ}$

Energy of the photons is

$\begin{aligned} & E=\frac{h c}{\lambda} \\ & E=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{0.45 \times 10^{-10}} \\ & E=4.413 \times 10^{-15} \mathrm{~J} \\ & E=27.6 \mathrm{keV}\end{aligned}$

b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?

Answer:

In such a tube where X-ray of energy 27.6 keV is to be produced the electrons should be having an energy about the same value and therefore accelerating voltage should be of order 30 KeV.

9. In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy $10.2\hspace{1mm}BeV$ into two $\gamma$ -rays of equal energy. What is the wavelength associated with each $\gamma$ -ray? ( $1\hspace{1mm}BeV=10^9\hspace{1mm}eV$ )

Answer:

The total energy of 2 $\gamma$ rays=10.2 BeV

The average energy of 1 $\gamma$ ray, E=5.1 BeV

The wavelength of the gamma-ray is given by

$\begin{aligned} & \lambda=\frac{c}{\nu} \\ & \lambda=\frac{h c}{h \nu} \\ & \lambda=\frac{h c}{E} \\ & \lambda=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{5.1 \times 10^9 \times 1.6 \times 10^{-19}} \\ & \lambda=2.436 \times 10^{-16} \mathrm{~m}\end{aligned}$

10.a) Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light. The number of photons emitted per second by a Medium wave transmitter of $10\hspace{1mm}kW$ power, emitting radiowaves of wavelength $500\hspace{1mm}m$ .

Answer:

The power emitted by the transmitter $(P)=10 \mathrm{~kW}$

Wavelengths of photons being emiited $=500 \mathrm{~m}$

The energy of one photon is $E$

$\begin{aligned}
E & =\frac{h c}{\lambda} \\
E & =\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{500} \\
E & =3.96 \times 10^{-28} \mathrm{~J}
\end{aligned}$

Number of photons emitted per second(n) is given by
$\begin{aligned}
n & =\frac{P}{E} \\
n & =\frac{10000}{3.96 \times 10^{-28}} \\
n & =2.525 \times 10^{31} \mathrm{~s}^{-1}
\end{aligned}$

10.b) Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light.The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive $\left(\sim 10^{-10} \mathrm{Wm}^2\right)$. Take the area of the pupil to be about 0.4 cm2, and the average frequency of white light to be about $6 \times 10^{14} \mathrm{~Hz}$.

Answer:

The minimum perceivable intensity of white light(I)=10 -10 Wm -2

Area of the pupil(A)=0.4 cm 2 =4 $\times$ 10 -5 m 2

Power of light falling on our eyes at minimum perceivable intensity is P

P=IA

P=10 -10 $\times$ 4 $\times$ 10 -5

P=4 $\times$ 10 -15 W

The average frequency of white light( $\nu$ )=6 $\times$ 10 14 Hz

The average energy of a photon in white light is

$\\E=h\nu$
$E=6.62\times 10^{-34}\times 6\times 10^{14}$
$E=3.972\times 10^{-19} J$

Number of photons reaching our eyes is n

$n=\frac{P}{E}$
$n=\frac{4\times 10^{-15}}{3.972\times 10^{-19}}$
$n=1.008\times 10^{4}s^{-1}$

11. Ultraviolet light of wavelength $2271\hspace{1mm}\dot{A}$ from a $100\hspace{1mm}W$ mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is $-1.3\hspace{1mm}V$, estimate the work function of the metal. How would the photo-cell respond to a high intensity $(\sim 10^5\hspace{1mm}Wm^2)$ red light of wavelength $6382\hspace{1mm}\dot{A}$ produced by a $He-Ne$ laser?

Answer:

The energy of the incident photons is E given by

$E=\frac{hc}{\lambda }$
$E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{2271\times 10^{-10}\times 1.6\times 10^{-19}}$
$E=5.465 eV$

Since stopping potential is -1.3 V work function is

$\phi _{0}=5.465-1.3$
$\phi _{0}=4.165 eV$

The energy of photons which red light consists of is E R

$E_{R}=\frac{hc}{\lambda _{R}}$
$E_{R}=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{6382\times 10^{-10}\times 1.6\times 10^{-19}}$
$ E_{R}=1.945eV$

Since the energy of the photons which red light consists of have less energy than the work function, there will be no photoelectric emission when they are incident.

12. Monochromatic radiation of wavelength $640.2 \mathrm{~nm}\left(1\mathrm{~nm}=10^{-9} \mathrm{~m}\right)$from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be $0.54\hspace{1mm}V$ . The source is replaced by an iron source and its $427.2\hspace{1mm}nm$ line irradiates the same photo-cell. Predict the new stopping voltage.

Answer:

The wavelength of photons emitted by the neon lamp $=640.2 \mathrm{~nm}$

The energy of photons emitted by the neon lamp is E given by
$\begin{aligned}
& E_1=\frac{h c}{\lambda} \\
& E_1=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{640.2 \times 10^{-9} \times 1.6 \times 10^{-19}} \\
& E_1=1.939 \mathrm{eV}
\end{aligned}$

Stopping potential is 0.54 V
Work function is therefore
$\begin{aligned}
\phi_0 & =1.939-0.54 \\
\phi_0 & =1.399 \mathrm{eV}
\end{aligned}$

The wavelength of photons emitted by the iron source $=427.2 \mathrm{~nm}$
The energy of photons emitted by the ion source is
$\begin{aligned}
& E_2=\frac{h c}{\lambda} \\
& E_2=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{427.2 \times 10^{-9} \times 1.6 \times 10^{-19}} \\
& E_2=2.905 \mathrm{eV}
\end{aligned}$

New stopping voltage is
$E_2-\phi_0=2.905-1.399=1.506 \mathrm{~V}$

13. A mercury lamp is a convenient source for studying the frequency dependence of photoelectric emission since it gives several spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:

$\lambda_1=3650 \dot{A}, \lambda_2=4047 \dot{A}, \lambda_3=4358 \dot{A}, \lambda_4=5461 \dot{A}, \lambda_5=6907 \dot{A}$

The stopping voltages, respectively, were measured to be

$V_{01}=1.28 \mathrm{~V}, V_{02}=0.95 \mathrm{~V}, V_{03}=0.74 \mathrm{~V}, V_{04}=0.16 \mathrm{~V}, V_{05}=0 \mathrm{~V}$

Determine the value of Planck's constant $h$, the threshold frequency and the work function for the material.

Answer:

$h\nu =\phi _{0}+eV$
$V=(\frac{h}{e})\nu -\phi_{0}\\$

where V is stopping potential, h is planks constant, e is electronic charge, $\nu$ is frequency of incident photons and $\phi _{0}$ is work function of metal in electron Volts.

To calculate the planks constant from the above date we plot the stopping potential vs frequency graph

$\nu_{1}=\frac{c}{\lambda_{1} }=\frac{3\times 10^{8}}{3650\times 10^{-10}}=8.219\times 10^{14}\ Hz$

$\nu_{2}=\frac{c}{\lambda_{2} }=\frac{3\times 10^{8}}{4047\times 10^{-10}}=7.412\times 10^{14}\ Hz$

$\nu_{3}=\frac{c}{\lambda_{3} }=\frac{3\times 10^{8}}{4358\times 10^{-10}}=6.884\times 10^{14}\ Hz$

$\nu_{4}=\frac{c}{\lambda_{4} }=\frac{3\times 10^{8}}{5461\times 10^{-10}}=5.493\times 10^{14}\ Hz$

$\nu_{5}=\frac{c}{\lambda_{5} }=\frac{3\times 10^{8}}{6907\times 10^{-10}}=4.343\times 10^{14}\ Hz$

The plot we get is

From the above figure, we can see that the curve is almost a straight line.

The slope of the above graph will give the Plank's constant divided by the electronic charge. The Planck's constant calculated from the above chart is

$h=\frac{\left ( 1.28-0.16 \right )\times 1.6\times 10^{-19}}{(8.214-5.493)\times 10^{14}}$
$h=6.573\times 10^{-34} Js$

Planks constant calculated from the above chart is therefore $6.573\times 10^{-34}\ Js$

14. The work function for the following metals is given: $Na:2.75\hspace{1mm}eV, K:2.30\hspace{1mm}eV, Mo:4.17\hspace{1mm}eV ,Ni:5.15\hspace{1mm}eV.$

Which of these metals will not give photoelectric emission for a radiation of wavelength $3300\hspace{1mm}\dot{A}$ from a $He-Cd$ laser placed $1\hspace{1mm}m$ away from the photocell? What happens if the laser is brought nearer and placed $50\hspace{1mm}cm$ away?

Answer:

The wavelength of the incident photons= $3300\dot{A}$

The energy of the incident photons is

$E=\frac{hc}{\lambda }$
$E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{3300\times 10^{-10}\times 1.6\times 10^{-19}}$
$ E=3.16 eV$

Mo and Ni will not give photoelectric emission for radiation of wavelength $3300\hspace{1mm}\dot{A}$ from a $He-Cd$ .

If the laser is brought nearer no change will be there in case of Mo and Ni although there will be more photoelectrons in case of Na and K.

15. Light of intensity $10^{-5} \mathrm{Wm}^{-2}$ falls on a sodium photo-cell of surface area $2\hspace{1mm}cm^2$ . Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about $2\hspace{1mm}eV$ . What is the implication of your answer? (Effective atomic area of a sodium atom = 10 -20 m 2 )

Answer:

Intensity of Incident light(I) = $10^{-5} \mathrm{Wm}^{-2}$

The surface area of the sodium photocell (A)=2 cm 2 = 2 $\times$ 10 -4 m 2

The rate at which energy falls on the photo cell=IA=2 $\times$ 10 - 9 W

The rate at which each of the 5 surfaces absorbs energy= IA/5=4 $\times$ 10 -10 W

Effective atomic area of a sodium atom (A')= 10 -20 m 2

The rate at which each sodium atom absorbs energy is R given by

$\begin{aligned} R & =\frac{I A}{5} \times \frac{A^{\prime}}{A} \\ R & =\frac{10^{-5} \times 10^{-20}}{5} \\ R & =2 \times 10^{-26} \mathrm{~J} / \mathrm{s}\end{aligned}$

The time required for photoelectric emission is

$\begin{aligned} t & =\frac{\phi_0}{R} \\ t & =\frac{2 \times 1.6 \times 10^{-19}}{2 \times 10^{-26}} \\ t & =1.6 \times 10^7 \mathrm{~s} \\ t & \approx 0.507 \text { years }\end{aligned}$

16. Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to $1 \dot{A}$, which is of the order of inter-atomic spacing in the lattice) $\left(m_e=9.11 \times 10^{-31} \mathrm{~kg}\right)$.

Answer:

According to De Broglie's equation

$p=\frac{h}{\lambda }$

The kinetic energy of an electron with De Broglie wavelength $1\hspace{1mm}\dot{A}$ is given by

$\begin{aligned} K & =\frac{p^2}{2 m_e} \\ K & =\frac{h^2}{\lambda^2 2 m_e} \\ K & =\frac{\left(6.62 \times 10^{-34}\right)^2}{2 \times 10^{-20} \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-19}} \\ K & =149.375 \mathrm{eV}\end{aligned}$

The kinetic energy of photon having wavelength $1\hspace{1mm}\dot{A}$ is

$\begin{aligned} & E=\frac{h c}{\lambda} \\ & E=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-10} \times 1.6 \times 10^{-19}} \\ & E=12.375 \mathrm{keV}\end{aligned}$

Therefore for the given wavelength, a photon has much higher energy than an electron.

17.a) Obtain the de Broglie wavelength of a neutron of kinetic energy $150\hspace{1mm}eV$ . As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain.$\left(m_e=9.11 \times 10^{-31} \mathrm{~kg}\right)$

Answer:

Kinetic energy of the neutron(K)=150eV

De Broglie wavelength associated with the neutron is

$\begin{aligned} & \lambda=\frac{h}{p} \\ & \lambda=\frac{h}{\sqrt{2 m_N K}} \\ & \lambda=\frac{6.62 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 150 \times 1.6 \times 10^{-19}}} \\ & \lambda=2.327 \times 10^{-12} \mathrm{~m}\end{aligned}$

Since an electron beam with the same energy has a wavelength much larger than the above-calculated wavelength of the neutron, a neutron beam of this energy is not suitable for crystal diffraction as the wavelength of the neutron is not of the order of the dimension of interatomic spacing.

b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature $(27\hspace{1mm}^{\circ}C )$ . Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.

Answer:

Absolute temperature = 273+27=300K

Boltzmann's Constant=1.38 $\times$ 10 -23 J/mol/K

The de Broglie wavelength associated with the neutron is

$\begin{aligned} & \lambda=\frac{h}{p} \\ & \lambda=\frac{h}{\sqrt{2 m_N K}} \\ & \lambda=\frac{h}{\sqrt{3 k T}} \\ & \lambda=\frac{6.62 \times 10^{-34}}{\sqrt{3 \times 1.38 \times 10^{-23} \times 300}} \\ & \lambda=1.446 \dot{A}\end{aligned}$

Since this wavelength is comparable to the order of interatomic spacing of a crystal it can be used for diffraction experiments. The neutron beam is to be thermalised so that its de Broglie wavelength attains a value such that it becomes suitable for the crystal diffraction experiments.

18. An electron microscope uses electrons accelerated by a voltage of $50\hspace{1mm}kV$ . Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?

Answer:

The potential difference through which electrons are accelerated(V)=50kV.

Kinetic energy(K) of the electrons would be eV where e is the electronic charge

The De Broglie wavelength associated with the electrons is

$\begin{aligned} & \lambda=\frac{h}{\sqrt{2 m_e K}} \\ & \lambda=\frac{6.62 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-19} \times 50000}} \\ & \lambda=5.467 \times 10^{-12} \mathrm{~m}\end{aligned}$

The wavelength of yellow light = 5.9 $\times$ 10 -7 m

The calculated De Broglie wavelength of the electron microscope is about 10 5 more than that of yellow light and since resolving power is inversely proportional to the wavelength the resolving power of electron microscope is roughly 10 5 times than that of an optical microscope.

19. The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10-15m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams.

(Rest mass energy of electron $=0.511\hspace{1mm}MeV$ .)

Answer:

Rest mass of the electron

$=mc^2=0.511MeV$

Momentum

$P=\frac{h}{\lambda}=\frac{6.63\times 10^{-34}}{10^{-15}}$

using the relativistic formula for energy

$E^2=(CP)^2+(mc^2)^2$

$=(3\times10^8 \times 6.63\times 10^{-19})^2+(0.511\times1.6\times10^{-19})^2$

$\approx 1.98\times10^{-10} J$

20. Find the typical de Broglie wavelength associated with a $He$ atom in helium gas at room temperature ( $27\hspace{1mm}^{\circ}C$ ) and 1 atm pressure, and compare it with the mean separation between two atoms under these conditions.

Answer:

The kinetic energy K of a He atom is given by

$K=\frac{3}{2}kT$

m He i.e. mass of one atom of He can be calculated as follows

$m_{He}=\frac{4\times 10^{-3}}{N_{A}} =\frac{4\times 10^{-3}}{6.023\times 10^{23}}=6.64\times 10^{-27} kg$ (N A is the Avogadro's Number)

De Broglie wavelength is given by

$\begin{aligned} & \lambda=\frac{h}{p} \\ & \lambda=\frac{h}{\sqrt{2 m_{H e} K}} \\ & \lambda=\frac{h}{\sqrt{3 m_{H e} k T}} \\ & \lambda=\frac{6.62 \times 10^{-34}}{\sqrt{3 \times 6.64 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300}} \\ & \lambda=7.27 \times 10^{-11} \mathrm{~m}\end{aligned}$

The mean separation between two atoms is given by the relation

$d=\left ( \frac{V}{N} \right )^{\frac{1}{3}}\\$

From the ideal gas equation we have

$\begin{aligned} & P V=n R T \\ & P V=\frac{N R T}{N_A} \\ & \frac{V}{N}=\frac{R T}{P N_A}\end{aligned}$

The mean separation is therefore

$\begin{aligned} & d=\left(\frac{R T}{P N_A}\right)^{\frac{1}{3}} \\ & d=\left(\frac{k T}{P}\right)^{\frac{1}{3}} \\ & d=\left(\frac{1.38 \times 10^{-23} \times 300}{1.01 \times 10^5}\right)^{\frac{1}{3}} \\ & d=3.35 \times 10^{-9} \mathrm{~m}\end{aligned}$

The mean separation is greater than the de Broglie wavelength.

21. Compute the typical de Broglie wavelength of an electron in a metal at $27\hspace{1mm}^{\circ}C$ and compare it with the mean separation between two electrons in a metal which is given to be about 2*10-10m.

Answer:

The de Broglie wavelength associated with the electrons is

$\begin{aligned} & \lambda=\frac{h}{\sqrt{3 m_e k T}} \\ & \lambda=\frac{6.62 \times 10^{-34}}{\sqrt{3 \times 9.1 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300}} \\ & \lambda=6.2 \times 10^{-9} \mathrm{~nm}\end{aligned}$

The de Broglie wavelength of the electrons is comparable to the mean separation between two electrons.

Answer the following questions:

22.a) Quarks inside protons and neutrons are thought to carry fractional charges $[(+2/3)e;(-1/3)e]$ . Why do they not show up in Millikan’s oil-drop experiment?

Answer:

Quarks are thought to be tight within a proton or neutron by forces which grow tough if one tries to pull them apart. That is event though fractional charges may exist in nature, the observable charges are still integral multiples of the charge of the electron

Answer the following questions:

b) What is so special about the combination $e/m$ ? Why do we not simply talk of $e$ and $m$ separately?

Answer:

The speed of a charged particle is given by the relations

$v=\sqrt{2K\left ( \frac{e}{m} \right )}$

or

$v=Br\left ( \frac{e}{m} \right )$

As we can see the speed depends on the ratio e/m it is of such huge importance.

Answer the following questions:

c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?

Answer:

At ordinary pressure due to a large number of collisions among themselves, the gases have no chance of reaching the electrodes while at very low pressure these collisions decrease exponentially and the gas molecules have a chance of reaching the respective electrodes and therefore are capable of conducting electricity.

d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?

Answer:

The work function is defined as the minimum energy below which an electron will never be ejected from the metal. But when photons with high energy are incident it is possible that electrons from different orbits get ejected and would, therefore, come out of the atom with different kinetic energies.

e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations:
$E=hv,p=\frac{h}{\lambda }$
But while the value of $\lambda$ is physically significant, the value of $v$ (and therefore, the value of the phase speed $v \lambda$ ) has no physical significance. Why?

Answer:

The absolute energy has no significance because of the reference point being arbitrary and thus the inclusion of an arbitrary constant rendering the value of $\nu\lambda$ and . $\nu$ to have no physical significance as such.

The group speed is defined as

$V_{G}=\frac{h}{\lambda m}$

Due to the significance of the group speed the absolute value of wavelength has physical significance.

Class 12 physics NCERT Chapter 11: Higher Order Thinking Skills (HOTS) Questions

Q.1 Light of wavelength 6200 Å falls on a metal having a photoelectric work function 2 eV . What is the value of stopping potential?

Answer:

Energy corresponding to $6200 Å=\frac{12375}{6200} \mathrm{ev}=1.996 \mathrm{ev}=2 \mathrm{ev}$
The electron is emitted with 2 ev kinetic energies. So, the stopping potential is 2 v .
So, the stopping potential is 2 V .

Q.2 The photoelectric threshold frequency of a metal is $\nu$. When light of frequency $4 \nu$ is incident on the metal, the maximum kinetic energy of the emitted photoelectrons is -

Answer:

$\begin{aligned} & \mathrm{E}=\phi+\mathrm{K}_{\max } \\ & \mathrm{K}_{\max }=4 \mathrm{hv}-\mathrm{h} v=3 \mathrm{~h} v\end{aligned}$

Q.3 The de-Broglie wavelength of a proton $\left(\right.$ mass $\left.=1.6 \times 10^{-27} \mathrm{~kg}\right)$ accelerated through a potential difference of 1 kV is
Answer:
$\begin{aligned} & \lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}} \\ & \therefore \lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times\left(1.6 \times 10^{-27}\right) \times\left(1.6 \times 10^{-19}\right) \times 1000}} \\ & \text { or } \lambda=\frac{6.6 \times 10^{-34} \times 10^{22}}{\sqrt{1.6 \sqrt{20}}}=0.9 \times 10^{-12} \mathrm{~m}\end{aligned}$

Q.4 The de-Broglie wavelength of a neutron at $927^{\circ} \mathrm{C}$ is $\lambda$. What will be its wavelength at $27^{\circ} \mathrm{C}$ ?
Answer:

the de-Broglie wavelength of a material particle at temperature $T$ is given by

$
\begin{aligned}
& \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mkT}}}, \text { where is Boltzmann's constant. } \\
& \Rightarrow \lambda \propto \frac{1}{\sqrt{T}} \\
& \therefore \frac{\lambda_2}{\lambda_1}=\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}} \\
& \text { or } \frac{\lambda_2}{\lambda_1}=\sqrt{\frac{1200}{300}}=2 \\
& \therefore \lambda_2=2 \lambda_1=2 \lambda
\end{aligned}
$


Q.5 What is de Broglie wavelength of the wave associated with an electron that has been accelerated through a potential difference of 50V?
Answer:

De - Broglie wavelength $(\lambda)$ is independent of charge
The gain of kinetic energy by an electron is eV

$
\begin{aligned}
& \frac{1}{2} m v^2=e V \\
& v=\sqrt{\frac{2 e V}{m}}=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 50}{9.11 \times 10^{-31}}} \\
& =4.19 \times 10^6 \mathrm{~m} / \mathrm{s} \\
& \therefore \lambda=\frac{h}{m v}=\frac{6.62 \times 10^{-34}}{9.1 \times 10^{-31} \times 4,19 \times 10^6} \\
& =1.74 \times 10^{-10} \mathrm{~m}
\end{aligned}
$

Approach to Solve Questions of Dual Nature of Matter and Radiation Class 12

Begin with Photoelectric Effect:

  • See how light can eject electrons from a metal. This indicates that light is particle-like (photons), not wave-like.

Know Einstein's Photoelectric Equation:

  • Apply the equation $E_k=h \nu-\phi$ (Kinetic energy = photon energy - work function). Most numericals are derived from this!
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JEE Main high scoring chapters and topics

As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE

Understand Work Function:

  • It's the minimum amount of energy required to eject an electron from a metal. Each metal has a unique work function.
JEE Main Highest Scoring Chapters & Topics
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Learn About Threshold Frequency:

  • If the frequency of light is less than a specific value, no electrons are emitted-whatever the brightness of the light!

De Broglie Wavelength:

  • Matter also behaves like waves, Use $\lambda=\frac{h}{m v}$ to find the wavelength of a traveling particle (like an electron).

Practice Graph Questions:

  • Learn graphs like kinetic energy vs frequency or stopping potential vs frequency these are regular exam questions.

Try NCERT and Previous Year Questions

  • Highlight conceptual clarity and practice in-text, exercise, and sample paper questions to improve your preparation

What Extra Should Students Study Beyond NCERT for JEE/NEET?

NCERT explains the core concepts of Dual Nature of Radiation and Matter clearly, including photoelectric effect and de Broglie wavelength. But for JEE, students should also focus on graph-based questions, deep application of Einstein’s photoelectric equation. Please check the below JEE topics for complete preparation.

JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

Class 12 Physics Chapter 11 NCERT Solutions: Important Formulas

  • Energy of photon(E): E=hv=hc
    Where: v= frequency of photon and = wavelength of photon

  • Momentum of photon: $p= h/\lambda$
    Where: h= planck’s constant and the wavelength of photon

  • Velocity of charged particle passing through a potential v:
    v= 2eVm
    Where: e= charge of electron and m is the mass of electron

  • Einstein’s Photoelectric Equation: Energy Quantum of Radiation $K_{\max }=h v-\phi_0$
    Where: Kmax is the maximum kinetic energy of the emitted electron and is the work function of metal

Dual Nature of Matter and Radiation Class 12-Topics

The Class 12 NCERT Physics chapter on the Dual Nature of Matter and Radiation covers the following topics:

NCERT Solutions for Class 12 Physics: Chapter-Wise

Also, check NCERT Books and NCERT Syllabus here:

NCERT solutions subject-wise

NCERT Exemplar Class 12 Solutions

Frequently Asked Questions (FAQs)

1. What is the weightage of chapter Dual Nature of Radiation and Matter for CBSE board exam?

Around 3 to 5 marks questions can be expected from the NCERT Class 12 chapter 11 for CBSE board exam according to the previous year papers. Students can also use the help of NCERT Exemplar Problems for Physics for more problems.

2. Which topics are important from NCERT Class 12 Physics chapter 11 solutions?

All the topics of the chapter are important. Understand all the graphs mentioned in the NCERT book. Also understand all the formulas listed in the chapter. The CBSE papers are based on the NCERT Syllabus. So all the concepts listed in the chapter are important.

3. Why is the Photoelectric Effect Important?

The photoelectric effect was crucial in establishing the particle theory of light. It challenged the classical wave theory and led to the development of quantum theory, which is essential for understanding many modern technologies like semiconductors, solar cells, and lasers.

4. What is Electron Emission?

Electron emission refers to the process by which electrons are released from the surface of a material. This can happen through different mechanisms such as thermal emission, photoelectric emission, or field emission.

5. What is the significance of this chapter in competitive exams?

This chapter is important for competitive exams like NEET and JEE Mains as it covers key concepts of quantum physics, which often form the basis of many questions. Understanding the photoelectric effect, Einstein’s equation and the Davisson-Germer experiment will help in solving problems related to wave-particle duality.

6. How can I prepare effectively for this chapter?

Understanding concepts is key. Focus on understanding the basic ideas behind wave-particle duality, photoelectric effect, and de Broglie wavelength. Practice solving problems, draw diagrams, and refer to the previous year’s questions. Regular revision and problem-solving will boost your confidence for exams.

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Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

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Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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