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NCERT Solutions for Class 12 Physics Chapter 11 - Dual Nature of Radiation and Matter
Ever wondered why automatic doors are able to sense your movements or why solar panels turn sunlight into electric power? Such modern technologies are founded on the concepts presented in Class 12 Physics Chapter 11 - Dual Nature of Radiation and Matter. The chapter presents key concepts like the photoelectric effect, the wave-particle duality, the photoelectric equation developed by Einstein and the hypothesis of de Broglie, and this aids students to comprehend the characteristics of light and microscopic particles, that is, light behaves like a wave and also like a particle at the same time.
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- Dual nature of radiation and matter NCERT Solutions: Download PDF
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NCERT Solutions for Class 12 Physics Chapter 11 - Dual Nature of Radiation and Matter are prepared by experienced subject experts according to the latest CBSE Class 12 Physics syllabus. These NCERT Solutions for Class 12 Physics Chapter 11 - Dual nature of radiation and matter also offer answers in a step-by-step manner to all NCERT questions in exercises, making complex theories and numerical problems easier to learn. These well-organised NCERT solutions are very valuable to CBSE board exam preparation and also to competitive exams such as JEE and NEET, where conceptual understanding and problem-solving accuracy are crucial.
Dual nature of radiation and matter NCERT Solutions: Download PDF
The Class 12 Physics Chapter 11 - Dual nature of radiation and matter question answers are presented in a step-by-step manner, containing the answers to all the problems present in the textbook. Concepts such as Huygens' principle, interference, diffraction, and polarisation, which are very important, are also covered by the Class 12 physics chapter 11 Dual nature of radiation and matters questions answers and hence made easier to comprehend, both in board examinations as well as in competitive exams like JEE/NEET. The NCERT Solutions for Class 12 Physics Chapter 11 PDF can also be downloaded to revise anytime and anywhere.Download PDF
Class 12 Physics Chapter 11 - Dual nature of radiation and matter - Exercise Solutions
Dual nature of radiation and matter class 12 question answers (exercise questions) offer well-explained solutions to all questions in the textbook. These NCERT Class 12 physics chapter 11 Dual nature of radiation and matters questions answers explain the concepts of photoelectric effect, wave particle duality and de Broglie wavelength in a simplified manner, which can be easily grasped in exams like the board exams and also JEE/NEET.
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Download EBookQ11.1 (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,
$\begin{aligned} & e V_0=h \nu \\ & \nu=\frac{e V_0}{h} \\ & \nu=\frac{1.6 \times 10^{-19} \times 30 \times 10^3}{6.62 \times 10^{-34}} \\ & \nu=7.25 \times 10^{18} \mathrm{~Hz}\end{aligned}$
Q11.1 (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, the 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 a 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 Planck's constant and e is the 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
$\begin{aligned}
& E=h \nu \\
& E=6.62 \times 10^{-34} \times 8.2 \times 10^{14} \\
& E=5.42 \times 10^{-19} \mathrm{~J}
\end{aligned}$
The 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
$\begin{aligned}
& E=h \nu \\
& E=6.62 \times 10^{-34} \times 7.21 \times 10^{14} \\
& E=4.77 \times 10^{-19} J
\end{aligned}$
The maximum kinetic energy of the ejected electrons is
$\begin{aligned}
& K E_{\max }=\frac{1}{2} m v^2 \\
& K E_{\max }=\frac{9.1 \times 10^{-31} \times\left(6 \times 10^5\right)^2}{22^2} \\
& K E_{\max }=1.64 \times 10^{-19} \mathrm{~J}
\end{aligned}$
The 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, the 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.
Class 12 Physics Chapter 11 - Dual nature of radiation and matter: Additional Questions
The Class 12 Physics Chapter 11 - Dual nature of radiation and matter is an introduction to one of the most interesting findings of modern physics - the dual nature of light and matter, i.e. they can act like waves and like particles at the same time. This theory fills in the gap between quantum and classical physics. The extra problems in this chapter enable the students to reinforce their knowledge on such concepts as the photoelectric effect, de Broglie wavelength and photoelectric equation as proposed by Einstein to ensure an in-depth comprehension of the major concepts of the exams and their real-life use.
Q1.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 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 equation below, p is the momentum
$K=\frac{p^{2}}{2m_{e}}$
$K=\frac{(6.62\times 10^{-25})^{2}}{2\times 9.1\times 10^{-31}}$
$K=2.41\times 10^{-19}\ J$
Q2.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 $p$
$\begin{aligned}
& p=\sqrt{2 m_N K} \\
& p=\sqrt{2 \times 1.675 \times 10^{-27} \times 6.21 \times 10^{-21}} \\
& p=4.56 \times 10^{-24} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}$
The associated De Broglie wavelength is
$\begin{aligned}
\lambda & =\frac{h}{p} \\
\lambda & =\frac{6.62 \times 10^{-34}}{4.56 \times 10^{-24}} \\
\lambda & =1.45 \times 10^{-10} \mathrm{~m}
\end{aligned}$
The 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
$\begin{aligned}
& p=\sqrt{2 m K} \\
& p=\sqrt{2 \times 4.65 \times 10^{-26} \times 6.21 \times 10^{-21}} \\
& p=2.4 \times 10^{-23} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}$
The 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 a Potential Difference V is K=eV where e is the electronic charge.
The 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/$m_e$ =1.366 $\times$ $10^{11}$ C/kg
Answer:
Using the same formula, we get the speed of electrons to be 1.88 $\times$ $10^{9}$ 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 a 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 travelling, 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 would not 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 a circular orbit, the electron's velocity must be perpendicular to the magnetic field.
The force due to the magnetic field is therefore F =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 $A^{\circ}$
The 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 an X-ray of energy 27.6 keV is to be produced, the electrons should have an energy about the same value, and therefore accelerating voltage should be of the 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 $E$
$\begin{aligned}
E & =\frac{h c}{\lambda} \\
E & =\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{500} \\
E & =3.96 \times 10^{-28} \mathrm{~J}
\end{aligned}$
The number of photons emitted per second(n) is given by
$\begin{aligned}
n & =\frac{P}{E} \\
n & =\frac{10000}{3.96 \times 10^{-28}} \\
n & =2.525 \times 10^{31} \mathrm{~s}^{-1}
\end{aligned}$
Q9.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 the 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
$\\E=h\nu$
$E=6.62\times 10^{-34}\times 6\times 10^{14}$
$E=3.972\times 10^{-19} J$
The 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 the 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 ER
$E_{R}=\frac{hc}{\lambda _{R}}$
$E_{R}=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{6382\times 10^{-10}\times 1.6\times 10^{-19}}$
$ E_{R}=1.945eV$
Since the energy of the photons which red light consists of has 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
$\begin{aligned}
& E_2=\frac{h c}{\lambda} \\
& E_2=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{427.2 \times 10^{-9} \times 1.6 \times 10^{-19}} \\
& E_2=2.905 \mathrm{eV}
\end{aligned}$
The 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 the stopping potential, h is Planck's constant, e is the electronic charge, $\nu$ is the frequency of incident photons and $\phi _{0}$ is the work function of the metal in electron Volts.
To calculate Planck's constant from the above data, 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 Planck's constant divided by the electronic charge. The Planck's constant calculated from the above chart is
$h=\frac{\left ( 1.28-0.16 \right )\times 1.6\times 10^{-19}}{(8.214-5.493)\times 10^{14}}$
$h=6.573\times 10^{-34} Js$
Planck's 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 given by R
$\begin{aligned} R & =\frac{I A}{5} \times \frac{A^{\prime}}{A} \\ R & =\frac{10^{-5} \times 10^{-20}}{5} \\ R & =2 \times 10^{-26} \mathrm{~J} / \mathrm{s}\end{aligned}$
The time required for photoelectric emission is
$\begin{aligned} t & =\frac{\phi_0}{R} \\ t & =\frac{2 \times 1.6 \times 10^{-19}}{2 \times 10^{-26}} \\ t & =1.6 \times 10^7 \mathrm{~s} \\ t & \approx 0.507 \text { years }\end{aligned}$
Q15. Crystal diffraction experiments can be performed using X-rays or electrons accelerated through an 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
The de Broglie wavelength associated with the neutron is
$\begin{aligned} & \lambda=\frac{h}{p} \\ & \lambda=\frac{h}{\sqrt{2 m_N K}} \\ & \lambda=\frac{6.62 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 150 \times 1.6 \times 10^{-19}}} \\ & \lambda=2.327 \times 10^{-12} \mathrm{~m}\end{aligned}$
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 105 times more than that of yellow light, and since resolving power is inversely proportional to the wavelength, the resolving power of the electron microscope is roughly 105 times 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 H,e 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)
The 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 tightly bound within a proton or neutron by forces which grow stronger if one tries to pull them apart. That is, even 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, which 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 renders 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 the wavelength has physical significance.
Class 12 Physics Chapter 11 - Dual nature of radiation and matter: Higher Order Thinking Skills (HOTS) Questions
The Class 12 Physics Chapter 11 – Dual Nature of Radiation and Matter HOTS (Higher Order Thinking Skills) questions are designed to challenge students to think beyond basic definitions and formulas. These questions test conceptual clarity on topics like photoelectric effect, wave-particle duality, and electron emission, encouraging analytical and application-based learning. Solving these HOTS questions helps students develop strong problem-solving skills and boosts their confidence for competitive exams like JEE and NEET.
Q.1 Light of wavelength 6200 Å falls on a metal having a photoelectric work function of 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 energy. So, the stopping potential is 2V.
So, the stopping potential is 2V.
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 the 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}
$
CBSE Class 12th Syllabus: Subjects & Chapters
Select your preferred subject to view the chapters
Dual nature of radiation and matter NCERT Solutions: Topics
The NCERT Solutions for Class 12 Physics Chapter 11 - Dual Nature of Radiation and Matter cover all the important topics that explain how light and matter exhibit both wave-like and particle-like behaviour. Among the important ones is the photoelectric effect, the equation of photoelectric emission found by Einstein, and the de Broglie wavelength. These topics form a sound background to study the contents of modern physics and how they have been used in the technology and research world.
Dual nature of radiation and matter NCERT Solutions: Important Formulas
Significant equations in Class 12 Physics NCERT Chapter 11 Dual nature of radiation and matter assist students to understand and find answers to numerical problems of photoelectric effect, kinetic energy of emitted electrons and De Broglie wavelength of particles. These equations are the mathematical foundation of the analysis of quantum mechanics and wave-particle duality. They are very important in passing board exams and competitive exams such as JEE and NEET.
1. Einstein's Photoelectric Equation:
$
h \nu=\phi+K_{\max }
$
Where $h \nu=$ energy of incident photon, $\phi=$ work function, $K_{\max }=$ maximum kinetic energy of emitted electron.
2. Kinetic Energy of Emitted Electron:
$
K_{\max }=e V_s
$
Where $V_s=$ stopping potential.
3. Threshold Frequency:
$
\nu_0=\frac{\phi}{h}
$
4. Threshold Wavelength:
$
\lambda_0=\frac{h c}{\phi}
$
5. de Broglie Wavelength (general form):
$
\lambda=\frac{h}{p}=\frac{h}{m v}
$
6. de Broglie Wavelength of Electron (in terms of kinetic energy):
$
\lambda=\frac{h}{\sqrt{2 m K}}
$
7. de Broglie Wavelength of Electron (in terms of accelerating potential V):
$
\lambda=\frac{h}{\sqrt{2 m e V}}
$
8. Work Function:
$
\phi=h \nu_0
$
How Can NCERT Solutions for Class 12 Physics Chapter 11 Help in Exam Preparation?
NCERT Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter Solutions are very useful among students who are preparing to take board and competitive exams such as JEE and NEET. These Dual nature of radiation and matter class 12 question answers are easy to understand as they break down tough concepts like the photoelectric effect, the equation by Einstein and the wavelength of de Broglie into easy-to-understand steps. The NCERT Dual Nature of Radiation and Matter Class 12 Solutions take students through theory and numerical problems, making sure that there is conceptual clarity and calculation accuracy.
As a result of these Class 12 physics Dual nature of radiation and matter question answers, the students will be able to grasp the proper technique of defining derivations, learn some useful formulas, and improve their problem-solving strategy. The step-by-step explanations in the Dual Nature of Radiation and Matter Class 12 NCERT Solutions are also useful in cultivating the confidence to write accurate answers on the exams. Also, having an abundance of practice questions and HOTS problems, the Dual Nature of Radiation and Matter Class 12 Questions and Answers will help the students to exercise the knowledge efficiently and get higher marks in the board and entrance exams.
Approach to Solve Questions of Class 12 Physics Chapter 11 - Dual Nature of Radiation and Matter
The approach to solving questions of Class 12 Physics Chapter 11 - Dual Nature of Radiation and Matter involves understanding both the theoretical and numerical aspects of wave-particle duality. Students need to pay attention to the duality of light and matter behaviour and exercise the corresponding formulae, such as the photoelectric equation by Einstein and the de Broglie wavelength. A good understanding of these concepts plays a part in deriving conceptual and numerical questions effectively on exams.
- Start by understanding the concept of wave-particle duality, how light behaves both as a wave and as a stream of particles called photons.
- Study the photoelectric effect carefully, including Einstein's explanation, threshold frequency, and how the stopping potential changes with light intensity and frequency.
- Solve numerical problems on the photoelectric equation of Einstein: E=hf−ϕ to strengthen your calculation skills.
- Revise the de Broglie hypothesis and learn how to derive and use the formula λ=h/mv for different particles like electrons and protons.
- Learn the working and use of electron emission processes, i.e., thermionic, field, and photoelectric emission.
- Practice past year and NCERT exemplar questions in order to determine patterns in commonly asked questions.
- Draw neat and labelled diagrams of experimental setups, like the photoelectric effect apparatus, to improve clarity in written answers.
- Pay more attention to units and conversions (particularly to energy and wavelength problems) to eliminate simple errors in calculations.
- Connect the concepts to practical applications like cathode ray tubes, solar cells and electron microscopes to memorise the concept better.
Importance of NCERT Solutions for Class 12 Physics Chapter 11: Dual Nature of Radiation and Matter
Class 12 Physics chapter 11 Dual Nature of Radiation and Matter solutions are highly relevant in obtaining knowledge of one of the most important concepts of modern physics, i.e. wave-particle duality of light and matter. The Class 12 Physics NCERT Chapter 11 Dual nature of radiation and matter, is a transition between classical and quantum concepts, and it is very much scoring when a person has a clear understanding of it.
- Helps students learn about the duality of light and matter, that is, a combination of the concepts of waves and particles.
- Describes important phenomena like the photoelectric effect, which cannot be described by classical physics.
- Develops a good conceptual clarity over the photoelectric equation of Einstein and other related experimental observations.
- Brings in the idea of de Broglie wavelength, which is the foundation of matter waves.
- Plays an important role in CBSE board exams, where theory-based and numerical questions are frequently asked.
- Highly relevant for competitive exams like JEE and NEET, especially conceptual questions from the photoelectric effect and matter waves.
- Improves problem-solving skills through step-by-step numerical solutions involving energy, frequency, and wavelength.
- Establishes a solid basis for the future chapters, including Atoms and Nuclei.
What Extra Should Students Study Beyond NCERT for JEE/NEET?
NCERT explains the core concepts of the Dual Nature of Radiation and Matter clearly, including the photoelectric effect and the de Broglie wavelength. But for JEE, students should also focus on graph-based questions and the deep application of Einstein’s photoelectric equation. Please check the JEE topics below for complete preparation.
NCERT Solutions for Class 12 Physics: Chapter-Wise
The NCERT Class 12 Physics Solutions (Chapter-Wise) contain step-by-step solutions to all textbook questions, which can help students understand the often complicated things with ease. These solutions are made according to the recent syllabus of CBSE, and they are also very crucial in making preparation for the board examination, as well as competitive examinations such as JEE and NEET.
NCERT Books and NCERT Syllabus
NCERT Solutions Subject-wise
NCERT Solutions subject-wise links provide easy access to chapter-wise answers for all major subjects in one place. These solutions help students understand concepts clearly, practise textbook questions effectively, and prepare confidently for exams
NCERT Exemplar Class 12 Solutions
NCERT Exemplar Solutions for Class 12 provide advanced and application-based questions to strengthen conceptual understanding. These subject-wise links help students practise higher-level problems and prepare effectively for board exams as well as competitive exams like JEE and NEET.
Frequently Asked Questions (FAQs)
Q: Where can I find the Dual Nature of Radiation and Matter NCERT Solutions PDF?
A:
You can download the Dual Nature of Radiation and Matter NCERT Solutions PDF from trusted educational sites like Careers360. These solutions cover all exercise questions in an easy, step-by-step format.
Q: What does Chapter 11 Physics Class 12 NCERT Solutions cover?
A:
Chapter 11 Physics Class 12 NCERT Solutions explain key topics like electron emission, photoelectric effect, Einstein’s equation, and de Broglie’s hypothesis. These are important for both board and competitive exams.
Q: What is wave-particle duality in dual nature of matter and radiation class 12?
A:
Wave-particle duality refers to the ability of particles of matter, such as electrons and photons, to exhibit both wave-like and particle-like behaviour. This means that they can display properties of both waves, such as diffraction and interference, and particles, such as quantization and the photoelectric effect.
Q: Are there Dual Nature of Radiation and Matter Class 12 numericals with solutions?
A:
Yes, many resources provide numericals from Dual Nature of Radiation and Matter Class 12 with detailed solutions. Practicing these helps in understanding concepts and scoring well in exams.
Q: What do you mean by photoelectric effect in class 12 physics chapter 11 ncert solutions?
A:
The photoelectric effect is a phenomenon in which electrons are emitted from a metal surface when light shines on it. The effect occurs because the light is made up of individual packets of energy, called photons, which transfer energy to electrons in the metal surface, causing them to be emitted.
Q: How can you apply NCERT Solution methods to solve advanced questions for NEET and JEE on the dual nature of radiation and matter?
A:
Begin with a clear statement of knowns and unknowns, use standard formulae as in board-level questions, but be ready to tackle multi-step problems. For competitive exams, pay special attention to units, magnitude orders, and conceptual traps (such as relativistic corrections at high energies or comparing energies of photons and electrons).
Q: Why is the Photoelectric Effect Important?
A:
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.
Q: How do you decide between using the wave or particle concept for solving problems on radiation and matter in the NCERT Solutions?
A:
Use the wave model for explaining interference, diffraction, and polarization, and the particle model for phenomena like the photoelectric effect and Compton effect. Always refer to the nature of the phenomenon in the question to choose the right approach.
Q: What is the significance of this chapter in competitive exams?
A:
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.
Q: How come the photoelectric effect cannot be explained using classical physics?
A:
Classical physics fails as it does not explain the existence of the threshold frequency and the instantaneous emission of electrons.
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Popular CBSE Class 12th Questions
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)
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Option 2)
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Option 4)
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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)
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Option 2)
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| Option 3)
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Option 4)
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A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
| Option 1)
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Option 2)
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| Option 3)
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Option 4)
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In the reaction,
| Option 1)
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Option 2)
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Option 4)
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How many moles of magnesium phosphate, 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 |
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