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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.
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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.
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$
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$
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.
Answer:
The electrons with the maximum kinetic energy of 0.34 eV will have the maximum speed
$\begin{aligned} & K E_{\max }=0.34 \mathrm{eV} \\ & K E_{\max }=5.44 \times 10^{-20} \mathrm{~J} \\ & v_{\max }=\sqrt{\frac{2 K E_{\max }}{m}} \\ & v_{\max }=\sqrt{\frac{2 \times 5.44 \times 10^{-20}}{9.1 \times 10^{-31}}} \\ & v_{\max }=3.44 \times 10^5 \mathrm{~ms}^{-1}\end{aligned}$
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
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 $1.046 \times 10^{-27} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$.
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}$
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}$
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.
$h=slope\times e$
$h=4.12\times10^{-15}\times1.6\times10^{-19}$
$h=6.59210^{-34} Js$
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
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.
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
$\begin{aligned}
& \nu_0=\frac{\phi_0}{h} \\
& \nu_0=4.728 \times 10^{14} \mathrm{~Hz}
\end{aligned}$
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
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
$\begin{aligned}
\lambda & =\frac{h}{p} \\
\lambda & =\frac{6.62 \times 10^{-34}}{40} \\
\lambda & =1.655 \times 10^{-35} \mathrm{~m}
\end{aligned}$
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
$\begin{aligned}
\lambda & =\frac{h}{p} \\
\lambda & =\frac{6.62 \times 10^{-34}}{0.06} \\
\lambda & =1.1 \times 10^{-32} \mathrm{~m}
\end{aligned}$
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
$\begin{aligned}
\lambda & =\frac{h}{p} \\
\lambda & =\frac{6.62 \times 10^{-34}}{2.2 \times 10^{-9}} \\
\lambda & =3.01 \times 10^{-25} \mathrm{~m}
\end{aligned}$
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.
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
$\begin{aligned}
p & =\frac{h}{\lambda} \\
p & =\frac{6.62 \times 10^{-34}}{10^{-9}} \\
p & =6.62 \times 10^{-25} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}$
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
$E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{10^{-9}}\\$
$E=1.86\times 10^{-16}\ J$
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
$\begin{aligned}
K & =\frac{p^2}{2 m} \\
K & =\frac{\left(4.728 \times 10^{-24}\right)^2}{2 \times 1.675 \times 10^{-27}} \\
K & =6.67 \times 10^{-21} J
\end{aligned}$
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 .
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.
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
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.
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
$\begin{aligned}
& F_b=\frac{m v^2}{r} \\
& e v B=\frac{m v^2}{r} \\
& r=\frac{m v}{e B} \\
& r=\frac{5.2 \times 10^6}{1.76 \times 10^{11} \times 1.3 \times 10^{-4}} \\
& r=0.227 m
\end{aligned}$
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
$r=\frac{m_{e}v}{eB\sqrt{1-\frac{v^{2}}{c^{2}}}}$
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
$\begin{aligned}
& \frac{m_e v^2}{r}=e v B \\
& \frac{m_e v}{r}=e B \\
& \frac{m_e}{r} \times \sqrt{\frac{2 e V}{m_e}}=e B \\
& \sqrt{\frac{e}{m_e}}=\frac{\sqrt{2 V}}{B r} \\
& \frac{e}{m_e}=\frac{2 V}{r^2 B^2} \\
& \frac{e}{m_e}=\frac{2 \times 100}{\left(2.83 \times 10^{-4}\right)^2 \times(0.12)^2} \\
& \frac{e}{m_e}=1.73 \times 10^{11} C~kg^{-1}
\end{aligned}$
Answer:
The wavelength of photons with maximum energy=0.45 $A^{\circ}$
Energy of the photons is
$\begin{aligned} & E=\frac{h c}{\lambda} \\ & E=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{0.45 \times 10^{-10}} \\ & E=4.413 \times 10^{-15} \mathrm{~J} \\ & E=27.6 \mathrm{keV}\end{aligned}$
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.
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
$\begin{aligned} & \lambda=\frac{c}{\nu} \\ & \lambda=\frac{h c}{h \nu} \\ & \lambda=\frac{h c}{E} \\ & \lambda=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{5.1 \times 10^9 \times 1.6 \times 10^{-19}} \\ & \lambda=2.436 \times 10^{-16} \mathrm{~m}\end{aligned}$
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
$n=\frac{P}{E}$
$n=\frac{4\times 10^{-15}}{3.972\times 10^{-19}}$
$n=1.008\times 10^{4}s^{-1}$
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.
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
$E_2-\phi_0=2.905-1.399=1.506 \mathrm{~V}$
The stopping voltages, respectively, were measured to be
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 $6.573\times 10^{-34}\ Js$
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.
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.
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.
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.
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.
(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
$E^2=(CP)^2+(mc^2)^2$
$=(3\times10^8 \times 6.63\times 10^{-19})^2+(0.511\times1.6\times10^{-19})^2$
$\approx 1.98\times10^{-10} J$
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.
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:
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:
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.
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.
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.
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}
$
Begin with Photoelectric Effect:
Know Einstein's Photoelectric Equation:
Understand Work Function:
Learn About Threshold Frequency:
De Broglie Wavelength:
Practice Graph Questions:
Try NCERT and Previous Year Questions
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.
As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters
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
The Class 12 NCERT Physics chapter on the Dual Nature of Matter and Radiation covers the following topics:
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.
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.
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.
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.
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.
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.
Admit Card Date:17 April,2025 - 17 May,2025
Exam Date:01 May,2025 - 08 May,2025
Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.
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If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.
Let me know if you need any other tips for your math prep. Good luck with your studies!
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As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters