Additional Questions - Chapter 7 Dual Nature of Radiation and Matter 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions
MultipleChoice Questions
Question 1.
The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy $3 \mathrm{eV}$ fall on it is $4 \mathrm{eV}$. The stopping potential, in volt, is
(a) 2
(b) 4
(c) 6
(d) 10
Answer:
(b) 4
Hint:
Stopping potential, $\mathrm{V}_0=\frac{K_{\max }}{e}=\frac{4 \mathrm{eV}}{e}=4 \mathrm{v}$
Question 2.
If an electron and proton are propagating in the form of waves having the same $\lambda$, it
implies that they have the same-
(a) energy
(b) momentum
(c) velocity
(d) angular momentum
Answer:
(b) momentum
Hint: Momentum, $\mathrm{p}=\frac{h}{\lambda}$
As both electron and proton have same $\lambda$, so they have the same momentum

Question 3.
An electron of mass $m$ and charge $e$ is accelerated from rest through a potential difference $\mathrm{V}$ in vacuum. Its final velocity will be-
(a) $\sqrt{\frac{2 e V}{m}}$
(b) $\sqrt{\frac{e V}{m}}$
(c) $\frac{e v}{2 m}$
(d) $\frac{e v}{m}$
Answer:
(a) $\sqrt{\frac{2 e V}{m}}$

Hint:
K.E. gained by an electron when accelerated through a potential difference V, $\frac{1}{2} \mathrm{mv}^2=\mathrm{eV}$ or $\mathrm{v}=\sqrt{\frac{2 e V}{m}}$
Question 4.
The work function of a substance is $4.0 \mathrm{eV}$. The longest wavelength of light that can cause photoelectron emission from this substance is approximately
(a) $540 \mathrm{~nm}$
(b) $400 \mathrm{~nm}$
(c) $310 \mathrm{~nm}$
(d) $220 \mathrm{~nm}$
Answer:
(c) $310 \mathrm{~nm}$
Hint:
$
\lambda_0=\frac{h c}{W}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4.0 \times 1.6 \times 10^{-19}}=\mathrm{m}=310 \times 10^{-9} \mathrm{~m}=310 \mathrm{~nm}
$
Question 5.
Sodium and copper have work function $2.3 \mathrm{eV}$ and $4.5 \mathrm{eV}$ respectively. Then, the ratio of their threshold wavelength is nearest to-
(a) $1: 2$
(b) $4: 1$
(c) $2: 1$
(d) $1: 4$
Answer:
(c) $2: 1$
Hint:
$
\frac{\lambda_0(\mathrm{Na})}{\lambda_0(\mathrm{Cu})}=\frac{\mathrm{W}_0(\mathrm{Cu})}{\mathrm{W}_0(\mathrm{Na})}
$

Question 6.
The surface of a metal is illuminated with the light of $400 \mathrm{~nm}$. The kinetic energy of the ejected photoelectrons was found to be $1.68 \mathrm{eV}$. The work function of the metal is $(\mathrm{hc}=$ $1240 \mathrm{eV} \mathrm{nm}$ )
(a) $3.09 \mathrm{eV}$
(b) $1.41 \mathrm{eV}$
(c) $1.51 \mathrm{eV}$
(d) $1.68 \mathrm{eV}$
Anwer:
(b) $1.41 \mathrm{eV}$
Hint:
$
\begin{aligned}
& \mathrm{K}_{\max }=\frac{h c}{\lambda} \mathrm{W}_0 \text { or } \mathrm{W}_0=\frac{h c}{\lambda}-\mathrm{K}_{\max } \\
& =\frac{1240}{400}-1.68=3.10-1.68=1.42 \mathrm{ev}
\end{aligned}
$
Question 7.
$4 \mathrm{eV}$ is the energy of the incident photon and the work function is $2 \mathrm{eV}$. The stopping potential will be
(a) $2 \mathrm{~V}$
(b) $4 \mathrm{~V}$
(c) $6 \mathrm{~V}$
(d) $2 \sqrt{ } 2 \mathrm{~V}$
Answer:
(d) $2 \sqrt{ } 2 \mathrm{~V}$
Hint:
$
\begin{aligned}
& \mathrm{eV}_0=\mathrm{hv}-\mathrm{W}_0=4 \mathrm{eV}-2 \mathrm{eV}=2 \mathrm{eV} \\
& \therefore \mathrm{V}_0=\frac{2 e v}{e}=2 \mathrm{v}
\end{aligned}
$
Question 8.
A light having wavelength $300 \mathrm{~nm}$ falls on a metal surface work function of metal is 2.54 $\mathrm{eV}$. What is stopping potential?
(a) $1.4 \mathrm{~V}$
(b) $2.59 \mathrm{~V}$
(c) $1.60 \mathrm{~V}$

(d) $1.29 \mathrm{~V}$
Answer:
(a) $1.4 \mathrm{~V}$
Hint:
$
\begin{aligned}
& \mathrm{eV}_0=\mathrm{hu}-\mathrm{W}_0=2 \mathrm{eV}-0.6 \mathrm{eV}=1.4 \mathrm{eV} \\
& \therefore \mathrm{V}_0=\frac{1.4 e V}{e}=1.4 \mathrm{eV}
\end{aligned}
$
Question 9.
If the kinetic energy of a free electron doubles, its de-Broglie wavelength changes by the factor
(a) $\frac{1}{2}$
(b) 2
(c) $\frac{1}{\sqrt{ } 2}$
(d) $\sqrt{ } 2$
Answer:
(c) $\frac{1}{\sqrt{ } 2}$
Hint:
$
\lambda=\frac{h}{\sqrt{2 m \mathrm{~K}}}
$
When kinetic energy is doubled, $\lambda^{\prime}=\frac{h}{\sqrt{2 m \times 2 K}}=\frac{1}{\sqrt{ } 2} \lambda$
Question 10.
If the kinetic energy of a particle is increased by 16 times, the percentage change in the de-Broglie wavelength of the particle is
(a) $25 \%$
(b) $75 \%$
(c) $60 \%$
(d) $50 \%$
Answer:
(b) $75 \%$
Hint:
$
\lambda=\frac{h}{\sqrt{2 m K}} ; \frac{h}{\sqrt{2 m \times 16 K}}=\frac{\lambda}{4}
$
$\%$ change in de-Broglie wavelength, $\frac{\lambda-\lambda^{\prime}}{\lambda}=\left[1-\frac{\lambda}{\lambda^{\prime}}\right] \times 100\left[1-\frac{1}{4}\right] \times 100=75 \%$

Question 11.
When a proton is accelerated through IV, then its kinetic energy will be
(a) $1 \mathrm{eV}$
(b) $13.6 \mathrm{eV}$
(c) $1840 \mathrm{eV}$
(d) $0.54 \mathrm{eV}$
Answer:
(a) $1 \mathrm{eV}$
Hint:
$
\mathrm{K}=\mathrm{qV}=\mathrm{e} \times 1 \mathrm{~V}=1 \mathrm{eV}
$
Question 12.
The kinetic energy of an electron, which is accelerated in the potential difference of 100 volts, is
(a) $416.6 \mathrm{cal}$
(b) $6.636 \mathrm{cal}$
(c) $1.602 \times 10^{-17} \mathrm{~J}$
(d) $1.6 \times 104 \mathrm{~J}$
Answer:
(c) $1.602 \times 10^{-17} \mathrm{~J}$
Hint:
$
\begin{aligned}
& \mathrm{K}=\mathrm{eV}=1.602 \times 10(\mathrm{c}) 1.602 \times 10^{-19} \times 100 \mathrm{~J} \\
& =1.602 \times 10(\mathrm{c}) 1.602 \times 10^{-17} \mathrm{~J}
\end{aligned}
$
Question 13.
Kinetic energy of emitted electron depends upon
(a) frequency
(b) intensity
(c) nature of atmosphere surrounding the electron
(d) none of these
Answer:
(a) frequency
Hint:

Kinetic energy of emitted electron depends on the frequency of incident radiation.
Question 14.
The work function of photometal is $6.626 \mathrm{eV}$. What is the threshold wavelength?
(a) $3921 \AA$
(b) $1875 Å$
(c) $1867 Å$
(d) $4433 Å$
Answer:
(b) $1875 Å$
Hint:
$
\lambda_0=\frac{h c}{W_0}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8 \times 10^{10}}{6.626 \times 1.6 \times 10^{-19}} \AA=1875 Å
$
Question 15.
The number of photo-electrons emitted for light of a frequency $v$ (higher than the
threshold frequency $v_0$ ) is proportional to
(a) Threshold frequency $\left(v_0\right)$
(b) Intensity of light
(c) Frequency of light (v)
(d) $v-v_0$
Answer:
(b) Intensity of light
Hint:
Photoelectric current oc Intensity of incident light
Question 16.
The speed of an electron having a wavelength of $10^{-10} \mathrm{~m}$ is
(a) $7.25 \times 10^6 \mathrm{~ms}^{-1}$
(b) $6.26 \times 10^6 \mathrm{~ms}^{-1}$
(c) $5.25 \times 10^6 \mathrm{~ms}^{-1}$
(d) $4.24 \times 10^6 \mathrm{~ms}^{-1}$
Answer:

(a) $7.25 \times 10^6 \mathrm{~ms}^{-1}$
Hint:
$
\begin{aligned}
& \text { As } \lambda=\frac{h}{m v} \\
& \therefore \mathrm{v}=\frac{h}{m \lambda}=\frac{6.6 \times 10^{-34}}{9.1 \times 10^{-31} \times 10^{-10}}=7.25 \times 10^6 \mathrm{~ms}^{-1}
\end{aligned}
$
Question 17.
If an electron and a photon propagate in the form of waves having the same wavelength, it implies that they have the same
(a) energy
(b) momentum
(c) angular momentum
(d) velocity
Answer:
(b) momentum
Hint:
As both electron and photon have same de-Broglie wavelength $(\lambda=\mathrm{h} / \mathrm{p})$, so they have the same momentum $P$.
Question 18.
Electron volt is a unit of
(a) Energy
(b) potential
(c) current
(d) charge
Answer:
(a) Energy
Hint:
Electron volt is a unit of energy

Question 19.
Photon of frequency $\mathrm{u}$ has a momentum associated with it. If $\mathrm{c}$ is the velocity of radiation, then the momentum is
(a) $\frac{h v}{c}$
(b) $\frac{v}{c}$
(c) huc
(d) $\frac{h}{c^2}$
Answer:
(a) $\frac{h v}{c}$
Hint:
$
\mathrm{P}=\frac{E}{c^2}=\frac{h v}{c}
$
Question 20.
The time taken by a photoelectron to come out after photon strikes is approximately
(a) $10^{-14} \mathrm{~s}$
(b) $10^{-10} \mathrm{~s}$
(c) $10^{-16} \mathrm{~s}$
(d) $10^{-1} \mathrm{~s}$
Answer:
(b) $10^{-10} \mathrm{~s}$
Hint:
The time lag between the incident of photon and the emission of photoelectrons is $10^{-10} \mathrm{~s}$ approximately.
Question 21.
Cathode rays consist of
(a) photons
(b) electrons
(c) protons
(d) $\alpha$-particles
Answer:
(b) electrons
Question 22.
The momentum of photon whose frequency is $f$ is

(a) $\frac{h f}{c}$
(b) $\frac{h c}{f}$
(c) $\frac{h}{f}$
(d) $\frac{c}{h f}$
Answer:
(a) $\frac{h f}{c}$
Hint:
$
\mathrm{p}=\mathrm{mc}=\frac{m c^2}{c}=\frac{h f}{c}
$
Question 23.
The energy of photon of wavelength $\lambda$ is
(a) $\frac{h c}{\lambda}$
(b) $\hat{h} \lambda \mathrm{c}$
(c) $\frac{\lambda}{h c}$
(d) $\frac{h \lambda}{c}$
Answer:
(a) $\frac{h c}{\lambda}$
Hint:
$
\mathrm{E}=\mathrm{h} v=\frac{h c}{\lambda}
$
Question 24.
The ratio of the energy of a photon with $\lambda=150 \mathrm{~nm}$ to that with $\lambda=300 \mathrm{~nm}$ is
(a) 2
(b) $\frac{1}{4}$
(c) 2
(d) $\frac{1}{2}$
Answer:
(a) 2
Hint:
$
\frac{E_1}{E_2}=\frac{\lambda_2}{\lambda_1}=\frac{300}{150}=2
$

Question 25.
Photons of $5.5 \mathrm{eV}$ energy fall on the surface of the metal emitting photoelectrons of maximum kinetic energy $4.0 \mathrm{eV}$. The stopping voltage required for these electrons is
(a) $5.5 \mathrm{~V}$
(b) $1.5 \mathrm{~V}$
(c) $9.5 \mathrm{~V}$
(d) $4.0 \mathrm{~V}$
Answer:
(d) $4.0 \mathrm{~V}$
Hint:
Stopping potential $=\frac{K_{\max }}{e}=\frac{4.0 e v}{e}=4.0 \mathrm{~V}$
Question 26.
The wavelength of photon is proportional to (where $v=$ frequency)
(a) $v$
(b) $\sqrt{v}$
(c) $\frac{1}{\sqrt{ } v}$
(d) $\frac{1}{v}$
Answer:
(d) $\frac{1}{v}$
Hint:
$\lambda=\frac{c}{v}$ i.e., $\lambda \propto \frac{1}{v}$
Question 27.
What is the energy of a photon whose wavelength is $6840 Å$ ?
(a) $1.81 \mathrm{eV}$
(b) $3.6 \mathrm{eV}$
(c) $-13.6 \mathrm{eV}$
(d) $12.1 \mathrm{eV}$
Answer:
(a) $1.81 \mathrm{eV}$
Hint:

$
\mathrm{E}=\mathrm{hv}=\frac{h c}{\lambda}=\frac{12400 e v \AA}{8840 Å}=1.81 \mathrm{eV}
$
Question 28.
Momentum of photon of wavelength $\lambda$ is
(a) $\frac{h v}{c}$
(b) zero
(c) $\frac{h \lambda}{c^2}$
(d) $\frac{\stackrel{h}{c}}{c}$
Answer:
(a) $\frac{h v}{c}$
Hint:
$
\mathrm{p}=\mathrm{mc}=\frac{m c^2}{c}=\frac{h v}{c}
$
Question 29.
The momentum of a photon of energy $1 \mathrm{MeV}$ in $\mathrm{kg} \mathrm{m} / \mathrm{s}$ will be
(a) $5 \times 10^{-22}$
(b) $0.33 \times 10^6$
(c) $7 \times 10^{-24}$
(d) $10^{-22}$
Answer:
(a) $5 \times 10^{-22}$
Hint:
$
\mathrm{P}=\frac{E}{c}=\frac{1 \mathrm{MeV}}{3 \times 10^8 \mathrm{~ms}^{-1}}=\frac{1.6 \times 10^{-13} \mathrm{~J}}{3 \times 10^8 \mathrm{~ms}^{-1}}=5.33 \times 10^{-22} \mathrm{Kg} \mathrm{ms}^{-1}
$

Question 30.
If we consider electrons and photons of same wavelength then will have same
(a) momentum
(b) angular momentum
(c) energy
(d) velocity
Answer:
(a) momentum
Hint:
As $\mathrm{p}=\mathrm{h} / \lambda$, so electrons and photons having the same wavelength $\lambda$ will have the same momentum p.
Question 31 .
Photoelectric effect can be explained by
(a) corpusular theory of light
(b) wave nature of light
(c) Bohr's theory
(d) quantum theory of light
Answer:
(d) quantum theory of light
Question 32 .
Which of the following waves can produce photoelectric effect?
(a) ultrasound
(b) infrared
(c) radiowaves
(d) X-rays
Answer:
(d) X-rays
Hint:
Electromagnetic radiation, being of high frequency such as X-rays can produce photoelectric effect.

Question 33.
Which light when falls on a metal will emit photoelectrons?
(a) uv radiation
(b) infrared radiation
(c) radio waves
(d) microwaves
Answer:
(a) uv radiation
Hint:
Ultraviolet radiation, being of high frequency, can emit photoelectrons from metals.
Question 34 .
In photoelectric effect, the $\mathrm{KE}$ of electrons emitted from the metal surface depends upon
(a) intensity of light
(b) frequency of incident light
(c) velocity of incident light
(d) both intensity and velocity of light
Answer:
(b) frequency of incident light
Hint:
The kinetic energy of photoelectrons depends upon the frequency of incident light.
Question 35 .
In photoelectric effect, electrons are ejected from metals, if the incident light has a certain minimum
(a) wavelength
(b) frequency
(c) amplitude
(d) angle of incidence
Answer:
(b) frequency
Hint:
For photoelectric emission, the incident light must have a certain minimum frequency, called threshold frequency.

Question 36.
Number of ejected photoelectrons increases with increases
(a) in intensity of light
(b) in wavelength of light
(c) in frequency of light
(d) never
Answer:
(a) in intensity of light
Hint:
Number of ejected photoelectrons increases with the increase in intensity of light.
Question 37.
By photoelectric effect, Einstein proved
(a) $E=h v$
(b) K.E. $=\frac{1}{2} \mathrm{mv}^2$
(c) $\mathrm{E}=\mathrm{mc}^2$
(d) $\mathrm{E}=\frac{-R h c^2}{n^2}$
Answer:
(a) $\mathrm{E}=\mathrm{h} v$
Hint:
Einstein explained photoelectric effect on the basis of planck's quantum theory of radiation and hence supported the relation : $E=h v$
Question 38.
A photocell employs photoelectric effect to convert
(a) change in the frequency of light into a change in the electric current
(b) Change in the frequency of light into a change in electric voltage
(c) Change in the intensity of illumination into a change in photoelectric current
(d) Change in the intensity of illumination into a change in the work function of the photo cathode
Answer:
(c) Change in the intensity of illumination into a change in photoelectric current

Hint:
It indicates that threshold frequency is greater than that of ultraviolet light. As X-rays have greater frequency than uv rays, so they can cause photoelectric effect.
Question 39.
When ultraviolet rays incident on metal plate there photoelectric effect does not occur, it occurs by incident of
(a) infrared rays
(b) X-rays
(c) radio waves
(d) microwave
Answer:
(b) X-rays
Hint:
It indicates that threshold frequency is greater than that of ultraviolet light. As X-rays have greater frequency than UV rays, so they can cause photoelectric effect.
Question 40.
The threshold frequency for photoelectric effect on sodiune corresponds to a wavelength of $5000 \AA$. Its function is
(a) $4 \times 10^{-19} \mathrm{~J}$
(b) $1 \mathrm{~J}$
(c) $2 \times 10^{-19} \mathrm{~J}$
(d) $3 \times 10^{-19} \mathrm{~J}$
Answer:
(a) $4 \times 10^{-19} \mathrm{~J}$
Hint:
$
\mathrm{W}_0=\frac{h c}{\lambda_0}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{5000 \times 10^{-10}} \mathrm{~J}=4 \times 10^{-19} \mathrm{~J}
$

Question 41.
The photoelectric work function for a metal surface is $4.125 \mathrm{eV}$. The cut off wavelength for this surface is
(a) $3000 Å$
(b) $2062.5 Å$
(c) $4125 Å$
(d) $6000 Å$
Answer:
(a) $3000 Å$
Hint:
$
\lambda_0=\frac{h c}{W_0}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4.125 \times 1.6 \times 10^{-19}} \mathrm{~m}=3 \times 10^{-7} \mathrm{~m}=3000 Å
$
Question 42 .
Ultraviolet radiations of $6.2 \mathrm{eV}$ falls on an aluminium surface. Kinetic energy of fastest electrons emitted is (work function $=4.2 \mathrm{eV}$ )
(a) $3.2 \times 10^{-21} \mathrm{~J}$
(b) $3.2 \times 10^{-19} \mathrm{~J}$
(c) $7 \times 10^{-25} \mathrm{~J}$
(d) $9 \times 10^{-32} \mathrm{~J}$
Answer:
(b) $3.2 \times 10^{-19} \mathrm{~J}$
Hint:
$
\begin{aligned}
& \mathrm{K}_{\max }=\mathrm{hv}-\mathrm{W}_0=6.2 \mathrm{eV}-4.2 \mathrm{eV} \\
& =2.0 \mathrm{eV}=2.0 \times 1.6 \times 10^{-19} \mathrm{~J}=3.2 \times 10^{-19} \mathrm{~J}
\end{aligned}
$
Question 43.
The de-Broglie wavelength of a tennis ball of mass $60 \mathrm{~g}$ moving with a velocity of $10 \mathrm{~ms}^{-}$¹ is approximately (planck's constant, $\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}$ )
(a) $10^{-33} \mathrm{~m}$
(b) $10^{-31} \mathrm{~m}$
(c) $10^{-16} \mathrm{~m}$
(d) $10^{-25} \mathrm{~m}$
Answer:
(a) $10^{-33} \mathrm{~m}$
Hint:
$
\lambda=\frac{h}{m v}=\frac{6.63 \times 10^{-34}}{60 \times 10^{-3} \times 10} \approx 10^{-33} \mathrm{~m}
$
Question 44.
The wavelength of de-Broglie wave is $2 \mu \mathrm{m}$, then its momentum $\left(\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}\right)$ is
(a) $3.315 \times 10^{-28} \mathrm{~kg} \mathrm{~ms}^{-1}$
(b) $1.66 \times 10^{-28} \mathrm{~kg} \mathrm{~ms}^{-1}$
(c) $4.97 \times 10^{-28} \mathrm{~kg} \mathrm{~ms}^{-1}$
(d) $9.9 \times 10^{-28} \mathrm{~kg} \mathrm{~ms}^{-1}$
Answer:
(a) $3.315 \times 10^{-28} \mathrm{~kg} \mathrm{~ms}^{-1}$
Hint:
$
\mathrm{p}=\frac{h}{\lambda}=\frac{6.03 \times 10^{-34} \mathrm{Js}}{2 \times 10^{-6} \mathrm{~m}}=3.315 \times 10^{-28} \mathrm{~kg} \mathrm{~ms}^{-1}
$

Question 45.
What is de-Broglie wavelength of electron having energy $10 \mathrm{KeV}$ ?
(a) $0.12 Å$
(b) $1.2 Å$
(c) $12.2 Å$
(d) none of these
Answer:
(a) $0.12 Å$
Hint:
$
\lambda=\frac{12.3}{\sqrt{ } v} Å=\frac{12.3}{\sqrt{10 \times 10^3}}=0.12 Å
$
Question 46.
Which one of the following property does not support wave theory of light?
(a) Light obeys laws of reflection and refraction
(b) Light waves get polarised
(c) Light shows photoelectric effect
(d) Light shows interference
Answer:
(c) Light shows photoelectric effect
Hint:
Photoelectric effect cannot be explained on the basis of wave theory of light.
Question 47.
de-Broglie wavelength $\lambda$ associated with neutrons is related with absolute temperature $\mathrm{T}$ as
(a) $\lambda \propto T$
(b) $\lambda \propto \frac{1}{T}$
(c) $\lambda \propto \frac{1}{\sqrt{ } T}$
(d) $\lambda \propto \mathrm{T}^2$
Answer:
(c) $\lambda \propto \frac{1}{\sqrt{ } T}$

Hint:
$
\lambda=\frac{h}{\sqrt{2 m K}}=\frac{h}{\sqrt{3 m K T}} \Rightarrow \lambda \propto \frac{1}{\sqrt{ } T}
$
Question 48.
As the intensity of incident light increases
(a) kinetic energy of emitted photoelectrons increases
(b) photoelectric current decreases
(c) photoelectric current increases
(d) kinetic energy of emitted photoelectrons decreases
Answer:
(c) photoelectric current increases
Hint:
As the intensity of incident light increases, photoelectric current increases.
Question 49.
The de Broglie wave corresponding to a particle of mass $\mathrm{m}$ and velocity $\mathrm{u}$ has a wavelength associated with it
(a) $\frac{h}{m v}$
(b) hmv
(c) $\frac{m h}{v}$
(d) $\frac{m}{h v}$
Answer:
(a) $\frac{h}{m v}$
Hint:
de-Broglie wavelength, $\lambda=\frac{h}{p}=\frac{h}{m v}$

Question 50.
If particles are moving with same velocity, then which has maximum de-broglie wavelength?
(a) Proton
(b) $\alpha$-particle
(c) Nevtron
(d) $\beta$-particle
Answer:
(d) $\beta$-particle
Hint:
As $\lambda=\mathrm{h} / \mathrm{mv}$, of the given particles $\beta$ - particle is the lightest, so it will have maximum de-Broglie wavelength.
Question 51 .
The dual nature of light is exhibited by
(a) diffraction and photoelectric effect
(b) photoelectric effect
(c) refraction and interference
(d) diffraction and reflection
Answer:
(a) diffraction and photoelectric effect
Hint:
Diffraction exhibits wave nature while photoelectric effect exhibits particle nature. Hence these two phenomena exhibit dual nature of light.
Question 52.
If the momentum of a particle is doubled, then its de-Broglie wavelength will-
(a) remain unchanged
(b) become four time
(c) become two times
(d) become half
Answer:
(d) become half
Hint:
As $\lambda=\frac{h}{p}$ when momentum $\mathrm{p}$ is doubled, wavelength will become half the initial value.

Question 53.
Moving with the same velocity, which of the following has the longest de-Broglie wavelength?
(a) $\beta$ - particle
(b) $\alpha$-particle
(c) proton
(d) neutron
Answer:
(a) $\beta$ - particle
Hint:
$
\lambda=\frac{h}{m v} \lambda \propto \frac{1}{m}
$
As $\beta$-particle (an electron) has the smallest mass, so it has the longest de-Broglie wavelength.
Question 54.
What is the de-Broglie wavelength of the a-particle accelerated through a potential difference of V volt? (mass of a-particle $=6.6455 \times 10^{-27} \mathrm{~kg}$ )
(a) $\frac{0.287}{\sqrt{ } V} Å$
(b) $\frac{12.27}{\sqrt{ } V} Å$
(c) $\frac{0.101}{\sqrt{ } V} Å$
(d) $\frac{0.202}{\sqrt{ } V} \AA$
Answer:
(c) $\frac{0.101}{\sqrt{ } V} Å$
Hint:
$\mathrm{K}=\mathrm{qV}=2 \mathrm{eV}$
$\lambda=\frac{h}{\sqrt{2 m K}}=\frac{h}{\sqrt{2 m \times 2 e V}}=\frac{h}{\sqrt{4 m e V}}$
$=\frac{6.63 \times 10^{-34}}{\sqrt{4 \times 6.6465 \times 10^{-27} \times 1.6 \times 10^{-19} \mathrm{~V}}} \mathrm{~m}=\frac{6.63 \times 10^{-11}}{6.52 \sqrt{\mathrm{V}}} \mathrm{m}=\frac{0.101}{\sqrt{\mathrm{V}}} Å$

Question 55.
A proton and an a - particle are accelerated through the same potential difference. The ratio of de-Broglie wavelength of proton to the de-Broglie wavelength of alpha particle will be
(a) $1: 2$
(b) $2 \sqrt{ } 2: 1$
(c) $2: 1$
(d) $1: 1$
Answer:
(b) $2 \sqrt{ } 2: 1$
Hint:
$
\begin{aligned}
& \lambda=\frac{h}{\sqrt{2 m q \mathrm{~V}}} \\
& \frac{\lambda_p}{\lambda_\alpha}=\sqrt{\frac{2 m_\alpha q_\alpha \mathrm{V}}{2 m_p q_p \mathrm{~V}}}=\sqrt{\frac{2 \times 4 m_p \times 2 e \times \mathrm{V}}{2 m_p \times e \times \mathrm{V}}}=2 \sqrt{2}: 1
\end{aligned}
$

Question 56.
Proton and $\alpha$-particle have the same de-Broglie wavelength. What is same for both of them?
(a) Time period
(b) Energy
(c) Frequency
(d) Momentum
Answer:
(d) Momentum
Hint:
$\lambda=\mathrm{h} / \mathrm{p}$, when wavelength $\lambda$ is same, momentump is also same.
Question 57.
The shortest wavelength of $\mathrm{X}$-ray emitted from an X-ray tube depends upon.
(a) the current in the tube
(b) the voltage applied to the tube
(c) the nature of the gas in the tube
(d) the atomic number of the target material
Answer:
(b) the voltage applied to the tube
Hint:
$$
\lambda_{\min }=\frac{h c}{e V} \text { i.e. }, \lambda_{\min } \propto \frac{1}{V}
$$
Question 58.
An $\mathrm{X}$-ray tube operates on $30 \mathrm{kV}$. The minimum wavelength emitted is $\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}$, $\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$.
(a) $6.6 Å$
(b) $0.133 Å$
(c) $1.2 Å$
(d) $0.4 Å$
Answer:
(d) $0.4 Å$

Hint:
$
\lambda_{\min }=\frac{h c}{e V}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 30 \times 10^3} \mathrm{~m}=0.4 \AA
$
Question 59.
The potential difference between the cathode and the target in a coolidge tube is $120 \mathrm{kV}$. What can be the minimum wavelength (in $Å$ ) of the X-rays emitted by this tube?
(a) $0.4 Å$
(b) $0.3 Å$
(c) $0.2 Å$
(d) $0.1 Å$
Answer:
(d) $0.1Å$
Hint:
$
\lambda_{\min }=\frac{12375}{V}=Å=\frac{12375}{120 \times 10^3} \AA=0.1 Å
$
Question 60 .
The work function for $\mathrm{Al}, \mathrm{K}$ and $\mathrm{Pt}$ is $4.28 \mathrm{eV}, 2.30 \mathrm{eV}$ and $5.65 \mathrm{eV}$ respectively. Their respective threshold frequencies would be
(a) $\mathrm{pt}>\mathrm{AL}>\mathrm{K}$
(b) $\mathrm{Al}>\mathrm{pt}>\mathrm{K}$
(c) $\mathrm{K}>\mathrm{AL}>\mathrm{pt}$
(d) $\mathrm{Al}>\mathrm{K}>\mathrm{pt}$
Answer:
(a) $\mathrm{pt}>\mathrm{AL}>\mathrm{K}$
Hint:
As $\mathrm{W}_0=\mathrm{hv}_0$ i.e., $\mathrm{W}_0 \propto \mathrm{v}_0$
$\mathrm{V}_0(\mathrm{pt})>>_0(\mathrm{AL})>\mathrm{V}_0(\mathrm{~K})$
Question 61 .
Among the following four spectral regions, the photons has the highest energy in
(a) Infrared
(b) Violet
(c) Red
(d) Blue
Answer:
(b) Violet

Hint:
$
\lambda_{\min }=\frac{h c}{e V}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 30 \times 10^3} \mathrm{~m}=0.4 Å
$
Question 59.
The potential difference between the cathode and the target in a coolidge tube is $120 \mathrm{kV}$. What can be the minimum wavelength (in $Å$ ) of the $\mathrm{X}$-rays emitted by this tube?
(a) $0.4 Å$
(b) $0.3 Å$
(c) $0.2 Å$
(d) $0.1 Å$
Answer:
(d) $0.1 Å$
Hint:
$
\lambda_{\min }=\frac{12375}{V}=\AA=\frac{12375}{120 \times 10^3} \AA=0.1 \AA
$
Question 60 .
The work function for $\mathrm{Al}, \mathrm{K}$ and $\mathrm{Pt}$ is $4.28 \mathrm{eV}, 2.30 \mathrm{eV}$ and $5.65 \mathrm{eV}$ respectively. Their respective threshold frequencies would be
(a) $\mathrm{pt}>\mathrm{AL}>\mathrm{K}$
(b) $\mathrm{Al}>\mathrm{pt}>\mathrm{K}$
(c) $\mathrm{K}>\mathrm{AL}>\mathrm{pt}$
(d) $\mathrm{Al}>\mathrm{K}>\mathrm{pt}$
Answer:
(a) $\mathrm{pt}>\mathrm{AL}>\mathrm{K}$
Hint:
As $\mathrm{W}_0=\mathrm{hv}_0$ i.e., $\mathrm{W}_0 \propto \mathrm{V}_0$
$\mathrm{V}_0(\mathrm{pt})>_0(\mathrm{AL})>\mathrm{V}_0(\mathrm{~K})$
Question 61.
Among the following four spectral regions, the photons has the highest energy in
(a) Infrared
(b) Violet
(c) Red
(d) Blue
Answer:
(b) Violet
Hint:

$\mathrm{E}=\frac{h c}{\lambda}$ Photon in violet region has least $\lambda$ and hence highest energy.
Short Answer Questions
Question 1.
Define electron volt. Express it value in joule.
Answer:
It is the kinetic energy gained by an electron when it is accelerated through a potential difference of 1 volt.
$
\begin{aligned}
& 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J} \\
& 1 \mathrm{MeV}=1.6 \times 10^{-13} \mathrm{~J}
\end{aligned}
$
Question 2.
What are photoelectrons?
Answer:
These are the electrons emitted from a metal surface when it is exposed to electro magnetic radiations of a suitable frequency.
Question 3 .
Define the term 'stopping potential' in relation to photoelectric effect.
Answer:
The minimum negative potential given to the anode of a photo-cell for which the photoelectric current becomes zero is called stopping potential.
Question 4.
Give some important uses of photo-cells.
Answer:
Applications of photo cells:

1. Photo cells have many applications, especially as switches and sensors.
2. Automatic lights that turn on when it gets dark use photocells, as well as street lights that switch on and off according to whether it is night or day.
3. Photo cells are used for reproduction of sound in motion pictures and are used as timers to measure the speeds of athletes during a race.
Question 5.
Why is a photo-cell also called an electric eye?
Answer:
Like an eye, a photo-cell can distinguish between a weak and an intense light. But a photocell gives a measure of light intensity in terms of photoelectric current. So it is also called an electric eye.
Question 6.
On what principle is an electron microscope based?
Answer:
As electron microscope exploits the wave nature of an accelerated beam of electrons (having a very small wavelength) to provide high magnifying and resolving powers.
Question 7.
What are X-ray spectra?
Answer:
$\mathrm{X}$-rays are produced when fast moving electrons strike the metal target. The intensity of the $\mathrm{X}$-rays when plotted against its wavelength gives a curve called $\mathrm{X}$-ray spectrum.
Long AnswerQuestions

Question 1.
Describe an experimental arrangement to study photoelectric effect.
Answer:
Experimental setup:
1. The apparatus is employed to study the phenomenon of photoelectric effect in detail .
2. $\mathrm{S}$ is a source of electromagnetic waves of known and variable frequency $\mathrm{v}$ and intensity I. C is the cathode (negative electrode) made up of photosensitive material and is used to emit electrons.

3. The anode (positive electrode) A collects the electrons emitted from $\mathrm{C}$. These electrodes are taken in an evacuated glass envelope with a quartz window that permits the passage of ultraviolet and visible light.
4. The necessary potential difference between $\mathrm{C}$ and $\mathrm{A}$ is provided by high tension battery $\mathrm{B}$ which is connected across a potential divider arrangement $\mathrm{PQ}$ through a key $\mathrm{K}$. $\mathrm{C}$ is connected to the centre terminal while A to the sliding contact $\mathrm{J}$ of the potential divider.
5. The plate A can be maintained at a desired positive or negative potential with respect to $\mathrm{C}$. To measure both positive and negative potential of $\mathrm{A}$ with respect to $\mathrm{C}$, the voltmeter is designed to have its zero marking at the centre and is connected between $\mathrm{A}$ and $\mathrm{C}$. The current is measured by a micro ammeter $\mu \mathrm{A}$ in series.
6. If there is no light falling on the cathode $\mathrm{C}$, no photoelectrons are emitted and the microammeter reads zero. When ultraviolet or visible light is allowed to fall on $\mathrm{C}$, the photoelectrons are liberated and are attracted towards anode.
7. As a result, the photoelectric current is setup in the circuit which is measured using micro ammeter.
8. The variation of photocurrent with respect to-
1. intensity of incident light
2. the potential difference between the electrodes
3. the nature of the material and
4. frequency of incident light can be studied with the help of this apparatus.

Question 2.
Write down the characteristics of photons.
Answer:
Characteristics of photons:
According to particle nature of light, photons are the basic constituents of any radiation and possess the following characteristic properties:
(i) The photons of light of frequency $\mathrm{v}$ and wavelength $\lambda$ will have energy, given by $\mathrm{E}=\mathrm{h} v=\frac{h c}{\lambda}$
(ii) The energy of a photon is determined by the frequency of the radiation and not by its intensity and the intensity has no relation with the energy of the individual photons in the beam.
(iii) The photons travel with the velocity of light and its momentum is given by $\mathrm{p}$
(iv) Since photons are electrically neutral, they are unaffected by electric and magnetic fields.
(v) When a photon interacts with matter (photon-electron collision), the total energy, total linear momentum and angular momentum are conserved. Since photon may be absorbed or a new photon may be produced in such interactions, the number of photons may not be conserved
Question 3.
Briefly explain the nature of light, (wave-particle duality)

Answer:
The nature of light: wave - particle duality We have learnt that wave nature of light explains phenomena such as interference, diffraction and polarization. Certain phenomena like black body radiation, photoelectric effect can be explained by assigning particle nature to light. Therefore, both theories have enough experimental evidences.
In the past, many scientific theories have been either revised or discarded when they contradicted with new experimental results. Here, two different theories are needed to answer the question: what is nature of light?
It is therefore concluded that light possesses dual nature, that of both particle and wave. It behaves like a wave at some circumstances and it behaves like a particle at some other circumstances.

In other words, light behaves as a wave during its propagation and behaves as a particle during its interaction with matter. Both theories are necessary for complete description of physical phenomena. Hence, the wave nature and quantum nature complement each other.
Question 4.
Derive de-Broglie wave equation (wavelength) for a material particle.
Answer:
De Broglie wave length:
The momentum of photon of frequency $\mathrm{v}$ is given by $\mathrm{p}=\frac{h v}{c}=\frac{h}{\lambda}$ since $\mathrm{c}=v \lambda$
The wavelength of a photon in terms of its momentum is $\lambda=\frac{h}{p} \ldots(1)$
According to de Broglie, the above equation is completely a general one and this is applicable to material particles as well. Therefore, for a particle of mass $\mathrm{m}$ travelling with speed $v$, the wavelength is given by $\lambda=\frac{h}{m v}=\frac{h}{p} \ldots .(2)$
This wavelength of the matter waves is known as de Broglie wavelength. This equation relates the wave character (the wave length $\lambda$ ) and the particle character (the momentum p) through Planck's constant.

Question 5.
Explain the production of X-rays.
Answer:
Production of $\mathrm{x}$-rays:
$\mathrm{X}$-rays are produced in x-ray tube which is essentially a discharge tube. A tungsten filament $\mathrm{F}$ is heated to incandescence by a battery. As a result, electrons are emitted from it by thermionic emission.

The electrons are accelerated to high speeds by the voltage applied between the filament $\mathrm{F}$ and the anode. The target materials like tungsten, molybdenum are embedded in the face of the solid copper anode. The face of the target is inclined at an angle with respect to the electron beam so that $x$-rays can leave the tube through its side.

When high-speed electrons strike the target, they are decelerated suddenly and lose their kinetic energy. As a result, x-ray photons are produced. Since most of the kinetic energy of the bombarding electrons gets converted into heat, targets made of high-meltmg-point metals and a cooling system are usually employed.
Question 6.
Briefly explain the concept of continuous X-ray spectra.
Answer:
Continuous x-ray spectra:
When a fast moving electron penetrates and approaches a target nucleus, the interaction between the electron and the nucleus either accelerates or decelerates it which results in a change of path of the electron. The radiation produced from such decelerating electron is called Bremsstrahlung or braking radiation

The energy of the photon emitted is equal to the loss of kinetic energy of the electron. Since an electron may lose part or all of its energy to the photon, the photons are emitted
with all possible energies (or frequencies). The continuous $\mathrm{x}$-ray spectrum is due to such radiations.
When an electron gives up all its energy, then the photon is emitted with highest frequency $v_0$ (or lowest wavelength $\lambda_0$ ). The initial kinetic energy of an electron is given by $\mathrm{eV}$ where $\mathrm{V}$ is the accelerating voltage. Therefore, we have $\mathrm{hv} v_0=\mathrm{eV}$ (or) $\frac{h c}{\lambda_0}=\mathrm{ev}$ $\lambda_0=\frac{h c}{e V}$
where $\lambda_0$ is the cut-off wavelength. Substituting the known values in the above equation, we get $\lambda_0=\frac{122400}{V} \AA$
The relation given by equation is known as the Duane - Hunt formula.
The value of $\lambda_0$ depends only on the accelerating potential and is same for all targets.
This is in good agreement with the experimental results. Thus, the production of continuous $\mathrm{x}$-ray spectrum and the origin of cut - off wavelength can be explained on the basis of photon theory of radiation.
Question 7.
Write down the applications of X-rays.
Answer:
Applications of $\mathrm{x}$-rays:
X-rays are being used in many fields. Let us list a few of them.
1. Medical diagnosis:
X-rays can pass through flesh more easily than through bones. Thus an x-ray radiograph containing a deep shadow of the bones and a light shadow of the flesh may be obtained. $\mathrm{X}$-ray radiographs are used to detect fractures, foreign bodies, diseased organs etc.
2. Medical therapy:
Since $\mathrm{x}$-rays can kill diseased tissues, they are employed to cure skin diseases, malignant tumours etc.

3. Industry:
$\mathrm{X}$-rays are used to check for flaws in welded joints, motor tyres, tennis balls and wood. At the custom post, they are used for detection of contraband goods.
4. Scientific research:
X-ray diffraction is important tool to study the structure of the crystalline materials - that is, the arrangement of atoms and molecules in crystals.