MCQ 511 Mark
AnswerCathode rays are beam of electrons.
View full question & answer→MCQ 521 Mark
For intensity I of a light of wavelength $5000 \mathring A$ the photoelectron saturation current is $0.40\ \mu \mathrm{A}$ and stopping potential is $1.36 \mathrm{~V}$, the work function of metal is
- A
$2.47 \mathrm{eV}$
- B
$1.36 \mathrm{eV}$
- ✓
$1.10 \mathrm{eV}$
- D
$0.43 \mathrm{eV}$
AnswerCorrect option: C. $1.10 \mathrm{eV}$
By using $E=W_0+K_{\max }$
$E=\frac{12375}{5000}=2.475 \mathrm{eV} \text { and } K_{\max }=e V_0=1.36 \mathrm{eV}$
So $2.475=W_0+1.36 \Rightarrow W_0=1.1 \mathrm{eV}$.
View full question & answer→MCQ 531 Mark
The work functions for sodium and copper are $2 \mathrm{eV}$ and $4 \mathrm{eV}$. Which of them is suitable for a photocell with $4000 \mathring A$ light
AnswerThreshold wavelength for $N a, \lambda_{N a}=\frac{12375}{2}=6187.5 \mathring A$
Also $\lambda_{C u}=\frac{12375}{4}=3093.75$
Since $\lambda_{N a}>4000 \mathring A$; So $N a$ is suitable.
View full question & answer→MCQ 541 Mark
If mean wavelength of light radiated by $100 W \operatorname{lamp}$ is $5000 A$, then number of photons radiated per second are
- A
$3 \times 10^{23}$
- B
$2.5 \times 10^{22}$
- ✓
$2.5 \times 10^{20}$
- D
$5 \times 10^{17}$
AnswerCorrect option: C. $2.5 \times 10^{20}$
$ P=\frac{n h c}{\lambda t} $
$\Rightarrow \frac{n}{t}=\frac{P . \lambda}{h c}=\frac{100 \times 5000 \times 10^{-10}}{6.6 \times 10^{-34} \times 3 \times 10^8} $
$ =2.50 \times 10^{20}$
View full question & answer→MCQ 551 Mark
What will be the ratio of de-Broglie wavelengths of proton and $\alpha$-particle of same energy
- ✓
$2: 1$
- B
$1: 2$
- C
$4: 1$
- D
$1: 4$
AnswerCorrect option: A. $2: 1$
$\lambda=\frac{h}{\sqrt{2 m E}}$
$ \Rightarrow \lambda \propto \frac{1}{\sqrt{m}}$
$ \Rightarrow \frac{\lambda_p}{\lambda_\alpha}=\sqrt{\frac{m_\alpha}{m_p}}=\frac{2}{1}$
View full question & answer→MCQ 561 Mark
The de-Broglie wavelength of a particle accelerated with 150 volt potential is $10^{-10} \mathrm{~m}$. If it is accelerated by $600$ volts p.d., its wavelength will be
- A
$0.25 \mathring A$
- ✓
$0.5 \mathring A$
- C
$1.5 \mathring A$
- D
$2 \mathring A$
AnswerCorrect option: B. $0.5 \mathring A$
$ \text { By using } \lambda \propto \frac{1}{\sqrt{V}} \Rightarrow \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{V_2}{V_1}}$
$ \Rightarrow \frac{10^{-10}}{\lambda_2}=\sqrt{\frac{600}{150}}=2 \Rightarrow \lambda=0.5 \mathring A .$
View full question & answer→MCQ 571 Mark
The frequency of the incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectrons is
Answer$ E=W_0+K_{\max } \Rightarrow K_{\max }=E-W_0=h v-W_0 $
$ \Rightarrow K_1=h v-W_0 \text { and } K_2=2 h v-W_0 \Rightarrow K_2>2 K_1$
View full question & answer→MCQ 581 Mark
The potential difference between the cathode and the target in a Collidge tube is $100\ \mathrm{kV}$. The minimum wavelength of the X-rays emitted by the tube is
- A
$0.66 \mathring A$
- B
$9.38 \mathring A$
- C
$0.246 \mathring A$
- ✓
$0.123 \mathring A$
AnswerCorrect option: D. $0.123 \mathring A$
$\lambda_{\min }=\frac{h c}{e V}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 100 \times 10^3}=0.123 \mathring A$
View full question & answer→MCQ 591 Mark
The velocity of photon is proportional to (where $v$ is frequency)
- A
$\frac{v^2}{2}$
- B
$\frac{1}{\sqrt{v}}$
- C
$\sqrt{v}$
- ✓
$v$
AnswerVelocity of photon $c=v \lambda$
View full question & answer→MCQ 601 Mark
The kinetic energy of an electron is $5 \mathrm{eV}$. Calculate the de-Broglie wavelength associated with it $\left(h=6.6 \times 10^{-} / \mathrm{s}, m=9.1 \times 10^{-} \mathrm{kg}\right)$
- ✓
$5.47 \mathring A$
- B
$10.9 \mathring A$
- C
$2.7 \mathring A$
- D
AnswerCorrect option: A. $5.47 \mathring A$
$ \lambda=\frac{h}{\sqrt{2 m E}}=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 5 \times 1.6 \times 10^{-19}}} $
$ =5.469 \times 10^{-10} \mathrm{~m}=5.47 \mathring A$
View full question & answer→MCQ 611 Mark
de-Broglie wavelength of a body of mass $1 \mathrm{~kg}$ moving with velocity of $2000 \mathrm{~m} / \mathrm{s}$ is
- ✓
$3.32 \times 10^{-27} \mathring A$
- B
$1.5 \times 10^{5} \mathring A$
- C
$0.55 \times 10^{-12} \mathring A$
- D
AnswerCorrect option: A. $3.32 \times 10^{-27} \mathring A$
$\lambda=\frac{h}{m v}=\frac{6.6 \times 10^{-34}}{1 \times 2000}=3.3 \times 10^{-37} \mathrm{~m}=3.3 \times 10^{-27} \mathring A$
View full question & answer→MCQ 621 Mark
$X$ rays cannot be deflected by means of an ordinary grating due to
View full question & answer→MCQ 631 Mark
The photoelectric effect can be understood on the basis of
- A
The principle of superposition
- B
The electromagnetic theory of light
- C
The special theory of relativity
- ✓
Line spectrum of the atom
AnswerCorrect option: D. Line spectrum of the atom
Photoelectric effect can be explained on the basis of spectrum of an atom.
View full question & answer→MCQ 641 Mark
An $\alpha$ particle is accelerated through a p.d of $10^6$ volt then K.E. of particle will be
- A
$8 \mathrm{MeV}$
- B
$4 \mathrm{MeV}$
- ✓
$2 \mathrm{MeV}$
- D
$1 \mathrm{MeV}$
AnswerCorrect option: C. $2 \mathrm{MeV}$
$K=Q . \Delta V=(2 e) \times 10^6 \mathrm{~V}=2 \times 10^6 \mathrm{eV}=2 \mathrm{MeV}$
View full question & answer→MCQ 651 Mark
A photon of energy $8 \mathrm{eV}$ is incident on a metal surface of threshold frequency $1.6 \times 10^{15} \mathrm{~Hz}$, then the maximum kinetic energy of photoelectrons emitted is $\left(h=6.6 \times 10^{-34} \mathrm{Js}\right)$
- A
$4.8 \mathrm{eV}$
- B
$2.4 \mathrm{eV}$
- ✓
$1.4 \mathrm{eV}$
- D
$0.8 \mathrm{eV}$
AnswerCorrect option: C. $1.4 \mathrm{eV}$
$\text { Work function } W_0=h v_0= 6.6 \times 10^{-34} \times 1.6 \times 10^{15} $
$ =1.056 \times 10^{-18} J=6.6 \mathrm{eV}$
$ \text { From } E=W_0+K_{\max } \Rightarrow K_{\max }=E-W_0=1.4 \mathrm{eV}$
From $E=W_0+K_{\max } \Rightarrow K_{\max }=E-W_0=1.4 \mathrm{eV}$
View full question & answer→MCQ 661 Mark
If the work function of a photometal is $6.825 \mathrm{eV}$. Its threshold wavelength will be $\left(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)$
- A
$1200 \mathring A$
- ✓
$1800 \mathring A$
- C
$2400 \mathring A$
- D
$3600 \mathring A$
AnswerCorrect option: B. $1800 \mathring A$
$\lambda_0=\frac{12375}{6.825}=1813 \mathring A \approx 1800 \mathring A$
View full question & answer→MCQ 671 Mark
An X-ray tube with a copper target emits $C u K_\alpha$ line of wavelength $1.50$ A. What should be the minimum voltage through which electrons are to be accelerated to produce this wavelength of $X$ rays
$\left(h=6.63 \times 10^{-34} \mathrm{~J} \text {-sec, } c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)$
- ✓
$8280 \mathrm{~V}$
- B
$828 \mathrm{~V}$
- C
$82800 \mathrm{~V}$
- D
$8.28 \mathrm{~V}$
AnswerCorrect option: A. $8280 \mathrm{~V}$
$ e V=\frac{h c}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.5 \times 10^{-10}} $
$ \Rightarrow V=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 1.5 \times 10^{-10}}=8280 \text { Volt. }$
View full question & answer→MCQ 681 Mark
Photon 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 are
- A
$5.5 \mathrm{~V}$
- B
$1.5 \mathrm{~V}$
- C
$9.5 \mathrm{~V}$
- ✓
$4.0 \mathrm{~V}$
AnswerCorrect option: D. $4.0 \mathrm{~V}$
$K_{\max }=e V_0 \Rightarrow e V_0=4 e V \Rightarrow V_0=4 \mathrm{~V}$
View full question & answer→MCQ 691 Mark
In Thomson experiment of finding $e / m$ for electrons, beam of electron is replaced by that of muons (particle with same charge as of electrons but mass $208$ times that of electrons). No deflection condition in this case satisfied if
- A
$B$ is increased $208$ times
- B
$E$ is increased $208$ times
- ✓
$B$ is increased $14.4$ times
- D
AnswerCorrect option: C. $B$ is increased $14.4$ times
In the condition of no deflection $\frac{e}{m}=\frac{E^2}{2 V B^2} \Rightarrow$ If $m$ is increased by $208$ times then $B$ should be increased $\sqrt{208}=14.4$ times
View full question & answer→MCQ 701 Mark
In an electron gun the control grid is given a negative potential relative to cathode in order to
- A
- ✓
Repel electrons and thus to control the number of electrons passing through it
- C
To select electrons of same velocity and to converge them along the axis
- D
To decrease the kinetic energy of electrons
AnswerCorrect option: B. Repel electrons and thus to control the number of electrons passing through it
View full question & answer→MCQ 711 Mark
A metal block is exposed to beams of $X$-ray of different wavelength. $X$-rays of which wavelength penetrate most
- ✓
$2 \mathring A$
- B
$4 \mathring A$
- C
$6 \mathring A$
- D
$8 \mathring A$
AnswerCorrect option: A. $2 \mathring A$
Penetrating power is greater for lower wavelength.
View full question & answer→MCQ 721 Mark
Hydrogen atom does not emit X-rays because
- ✓
Its energy levels are too close to each other
- B
Its energy levels are too apart
- C
- D
AnswerCorrect option: A. Its energy levels are too close to each other
View full question & answer→MCQ 731 Mark
X-rays were discovered by
MCQ 741 Mark
$X$-rays and gamma rays are both electromagnetic waves. Which of the following statements is true
- ✓
In general $X$-rays have larger wavelength than of gamma rays
- B
$X$-rays have smaller wavelength than that of gamma rays
- C
Gamma rays have smaller frequency than that of $X$-rays
- D
Wavelength and frequency of $X$-rays are both larger than that of gamma rays
AnswerCorrect option: A. In general $X$-rays have larger wavelength than of gamma rays
View full question & answer→MCQ 751 Mark
If $V$ be the accelerating voltage, then the maximum frequency of continuous $X$-rays is given by
- A
$\frac{e h}{V}$
- B
$\frac{h V}{e}$
- ✓
$\frac{e V}{h}$
- D
$\frac{h}{e V}$
AnswerCorrect option: C. $\frac{e V}{h}$
$E=e V=h v_{\max } \Rightarrow v_{\max }=\frac{e V}{h}$
View full question & answer→MCQ 761 Mark
The maximum velocity of an electron emitted by light of wavelength $\lambda$ incident on the surface of a metal of work function $\phi$, is Where $h=$ Planck's constant, $m=$ mass of electron and $c=$ speed of light.
- A
$\left[\frac{2(h c+\lambda \phi)}{m \lambda}\right]^{1 / 2}$
- B
$\frac{2(h c-\lambda \phi)}{m}$
- ✓
$\left[\frac{2(h c-\lambda \phi)}{m \lambda}\right]^{1 / 2}$
- D
$\left[\frac{2(h \lambda-\phi)}{m}\right]^{1 / 2}$
AnswerCorrect option: C. $\left[\frac{2(h c-\lambda \phi)}{m \lambda}\right]^{1 / 2}$
According to Einstein's photoelectric equation
$\frac{h c}{\lambda}=\phi+\frac{1}{2} m v^2 \Rightarrow v=\left[\frac{2(h c-\lambda \phi)}{m \lambda}\right]^{1 / 2}$
View full question & answer→MCQ 771 Mark
The specific charge of an electron is
- A
$1.6 \times 10^{-19}$ coulomb
- B
$4.8 \times 10^{-10}$ statcoulomb
- ✓
$1.76 \times 10^{11}$ coulomb $/ \mathrm{kg}$
- D
$1.76 \times 10^{-11}$ coulomb $/ \mathrm{kg}$
AnswerCorrect option: C. $1.76 \times 10^{11}$ coulomb $/ \mathrm{kg}$
$\frac{e}{m}=\frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}}=1.76 \times 10^{11} \mathrm{C} / \mathrm{kg}$
View full question & answer→MCQ 781 Mark
The retarding potential for having zero photo electron current
- A
Is proportional to the wavelength of incident light
- B
Increases uniformly with the increase in the wavelength of incident light
- C
Is proportional to the frequency of incident light
- ✓
Increases uniformly with the increase in the frequency of incident light wave
AnswerCorrect option: D. Increases uniformly with the increase in the frequency of incident light wave
Increases uniformly with the increase in the frequency of incident light wave
View full question & answer→MCQ 791 Mark
light of wavelength $1824 \mathring A$, incident on the surface of a metal, produces photo-electrons with maximum energy $5.3 \mathrm{eV}$. When light of wavelength $1216 \mathring A$ is used, the maximum energy of photoelectrons is $8.7 \mathrm{eV}$. The work function of the metal surface is
- A
$3.5 \mathrm{eV}$
- B
$13.6 \mathrm{eV}$
- C
$6.8 \mathrm{eV}$
- ✓
$1.5 \mathrm{eV}$
AnswerCorrect option: D. $1.5 \mathrm{eV}$
$E=W_0+K_{\max }$. From the given data $E$ is $6.78 \mathrm{eV}$
(for $\lambda=$ $1824 A$ ) or $10.17 \mathrm{eV}$ (for $\lambda=1216 \mathring A$ )
$\therefore W_0=E-K_{\max }=6.78-5.3=1.48 \mathrm{eV}$or$W_0=10.17-8.7=1.47 \mathrm{eV} \text {. }$
View full question & answer→MCQ 801 Mark
A cathode emits $1.8 \times 10^{14}$ electrons per second, when heated. When $400 \mathrm{~V}$ is applied to anode all the emitted electrons reach the anode. The charge on electron is $1.6 \times 10^{-19} \mathrm{C}$. The maximum anode current is
- A
$2.7 \mu \mathrm{A}$
- ✓
$29 \mu A$
- C
$72 \mu \mathrm{A}$
- D
$29 m A$
AnswerCorrect option: B. $29 \mu A$
$ i=\frac{Q}{t}=\frac{n e}{t}=1.8 \times 10^{14} \times 1.6 \times 10^{-19}=28.8 \times 10^{-6} \mathrm{~A}$
$ =29 \mu \mathrm{A}$
View full question & answer→MCQ 811 Mark
What is the difference between soft and hard $X$-rays
AnswerFrequency of hard $\mathrm{X}$-rays is greater than that of soft $\mathrm{X}$-rays.
View full question & answer→MCQ 821 Mark
The threshold wavelength for photoelectric effect of a metal is $6500\ \mathring A$. The work function of the metal is approximately
- ✓
$2 \mathrm{eV}$
- B
$1 \mathrm{eV}$
- C
$0.1 \mathrm{eV}$
- D
$3 \mathrm{eV}$
AnswerCorrect option: A. $2 \mathrm{eV}$
$\lambda_0=\frac{12375}{6500}=1.9 \mathrm{eV} \approx 2 \mathrm{eV}$.
View full question & answer→MCQ 831 Mark
The mass of a photo electron is
- A
$9.1 \times 10^{-27} \mathrm{~kg}$
- B
$9.1 \times 10^{-29} \mathrm{~kg}$
- ✓
$9.1 \times 10^{-31} \mathrm{~kg}$
- D
$9.1 \times 10^{-34} \mathrm{~kg}$
AnswerCorrect option: C. $9.1 \times 10^{-31} \mathrm{~kg}$
View full question & answer→MCQ 841 Mark
Electron volt is a unit of
MCQ 851 Mark
$K_\alpha$ characteristic $X$-ray refers to the transition
- ✓
$n=2$ to $n=1$
- B
$n=3$ to $n=2$
- C
$n=3$ to $n=1$
- D
$n=4$ to $n=2$
AnswerCorrect option: A. $n=2$ to $n=1$
View full question & answer→MCQ 861 Mark
If in a photoelectric experiment, the wavelength of incident radiation is reduced from $6000 \mathring A$ to $4000 \mathring A$ then
- A
Stopping potential will decrease
- ✓
Stopping potential will increase
- C
Kinetic energy of emitted electrons will decrease
- D
The value of work function will decrease
AnswerCorrect option: B. Stopping potential will increase
Stopping potential $V_0=\frac{h c}{e}\left[\frac{1}{\lambda}-\frac{1}{\lambda_0}\right]$.
As $\lambda$ decreases so $V_0$ increases.
View full question & answer→MCQ 871 Mark
An electron is moving with constant velocity along $x$-axis. If a uniform electric field is applied along $y$-axis, then its path in the $x-y$ plane will be
View full question & answer→MCQ 881 Mark
When cathode rays strike a metal target of high melting point with very high velocity, then
- ✓
$X$-rays are produced
- B
- C
- D
Ultrasonic waves are produced
AnswerCorrect option: A. $X$-rays are produced
View full question & answer→MCQ 891 Mark
Penetrating power of $X$-rays can be increased by
- ✓
Increasing the potential difference between anode and cathode
- B
Decreasing the potential difference between anode and cathode
- C
Increasing the cathode filament current
- D
Decreasing the cathode filament current
AnswerCorrect option: A. Increasing the potential difference between anode and cathode
With the increase in potential difference between anode and cathode energy of striking electrons increases which in turn increases the energy (penetration power) of X-rays.
View full question & answer→MCQ 901 Mark
In Thomson's method of determining $e / m$ of electrons
- A
Electric and magnetic fields are parallel to electrons beam
- ✓
Electric and magnetic fields are perpendicular to each other and perpendicular to electrons beam
- C
Magnetic field is parallel to the electrons beam
- D
Electric field is parallel to the electrons beam
AnswerCorrect option: B. Electric and magnetic fields are perpendicular to each other and perpendicular to electrons beam
View full question & answer→MCQ 911 Mark
When an inert gas is filled in the place vacuum in a photo cell, then
- A
Photo-electric current is decreased
- ✓
Photo-electric current is increased
- C
Photo-electric current remains the same
- D
Decrease or increase in photo-electric current does not depend upon the gas filled
AnswerCorrect option: B. Photo-electric current is increased
In the presence of inert gas photoelectrons emitted by cathode ionise the gas by collision and hence the current increases.
View full question & answer→MCQ 921 Mark
Which one of the following is true in photoelectric emission
- A
Photoelectric current is directly proportional to the amplitude of light of a given frequency
- ✓
Photoelectric current is directly proportional to the intensity of light of a given frequency at moderate intensities
- C
Above the threshold frequency, the maximum K.E. of photoelectrons is inversely proportional to the frequency of incident light
- D
The threshold frequency depends upon the wavelength of incident light
AnswerCorrect option: B. Photoelectric current is directly proportional to the intensity of light of a given frequency at moderate intensities
View full question & answer→MCQ 931 Mark
Light of wavelength $4000 \mathring A$ falls on a photosensitive metal and a negative $2\ V$ potential stops the emitted electrons. The work function of the material (in $\mathrm{eV}$ ) is approximately
$\left(h=6.6 \times 10^{-34} J s, e=1.6 \times 10^{-19} C, c=3 \times 10^8 \mathrm{~ms}^{-1}\right)$
- ✓
$1.1$
- B
$ 2.0$
- C
$2.2$
- D
$ 3.1$
AnswerEnergy of incident light $E(\mathrm{eV})=\frac{12375}{4000}=3.09 \mathrm{eV}$
Stopping potential is $-2 \mathrm{~V}$ so $K_{\max }=2 \mathrm{eV}$
Hence by using $E=W_0+K_{\max } ; W=1.09 \mathrm{eV} \approx 1.1 \mathrm{eV}$
View full question & answer→MCQ 941 Mark
An $X$-ray tube operates on $30 \mathrm{kV}$. What is the minimum wavelength emitted $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}, e=1.6 \times 10^{-19}\right.$ Coulomb,$\left.c=3 \times 10^8 \mathrm{~ms}\right)$
- A
$0.133 \mathring A$
- ✓
$0.4 \mathring A$
- C
$1.2 \mathring A$
- D
$6.6 \mathring A$
AnswerCorrect option: B. $0.4 \mathring A$
$\lambda_{\min }=\frac{12375}{V}=\frac{12375}{30 \times 10^3}=0.4 \mathring A$
View full question & answer→MCQ 951 Mark
Mosley measured the frequency $(f)$ of the characteristic $X$-rays from many metals of different atomic number $(Z)$ and represented his results by a relation known as Mosley's law. This law is (a, b are constants)
- ✓
$f=a(Z-b)^2$
- B
$Z=a(f-b)^2$
- C
$f^2=a(Z-b)$
- D
$f=a(Z-b)^{1 / 2}$
AnswerCorrect option: A. $f=a(Z-b)^2$
Mosley's law is $f=a(Z-b)^2$
View full question & answer→MCQ 961 Mark
For the production of $X$-rays of wavelength $0.1 \mathring A$ the minimum potential difference will be
- A
$12.4 \mathrm{kV}$
- B
$24.8 \mathrm{kV}$
- ✓
$124 \mathrm{kV}$
- D
$248 \mathrm{kV}$
AnswerCorrect option: C. $124 \mathrm{kV}$
$ \lambda_{\min }=\frac{h c}{e V}=\frac{12375}{V} A ; $
$\therefore V=\frac{12375}{\lambda \operatorname{in} A}=124 \mathrm{kV}$
View full question & answer→MCQ 971 Mark
In Millikan oil drop experiment, a charged drop of mass $1.8 \times 10^{-14} \mathrm{~kg}$ is stationary between its plates. The distance between its plates is $0.90 \mathrm{~cm}$ and potential difference is $2.0 \mathrm{kilo}$ volts. The number of electrons on the drop is
Answer$ Q E=m g \Rightarrow Q=\frac{m g}{E} \Rightarrow n=\frac{m g d}{V e} $
$\Rightarrow n=\frac{1.8 \times 10^{-14} \times 10 \times 0.9 \times 10^{-2}}{2 \times 10^3 \times 1.6 \times 10^{-19}}=5$
View full question & answer→MCQ 981 Mark
Penetrating power of $X$-rays depends on
- A
Current flowing in the filament
- ✓
Applied potential difference
- C
- D
AnswerCorrect option: B. Applied potential difference
The potential difference across the filament and target determines the energy and thence the penetrating power of $X$ rays.
View full question & answer→MCQ 991 Mark
When the light source is kept $20 \mathrm{~cm}$ away from a photo cell, stopping potential $0.6 \mathrm{~V}$ is obtained. When source is kept $40 \mathrm{~cm}$ away, the stopping potential will be
- A
$0.3 \mathrm{~V}$
- ✓
$0.6 \mathrm{~V}$
- C
$1.2 \mathrm{~V}$
- D
$2.4 \mathrm{~V}$
AnswerCorrect option: B. $0.6 \mathrm{~V}$
Stopping potential does not depend on the relative distance between the source and the cell.
View full question & answer→MCQ 1001 Mark
The maximum wavelength of radiation that can produce photoelectric effect in a certain metal is $200 \mathrm{~nm}$. The maximum kinetic energy acquired by electron due to radiation of wavelength $100 \mathrm{~nm}$ will be
- A
$12.4 \mathrm{eV}$
- ✓
$6.2 \mathrm{eV}$
- C
$100 \mathrm{eV}$
- D
$200 \mathrm{eV}$
AnswerCorrect option: B. $6.2 \mathrm{eV}$
$ K_{\max }(\mathrm{eV})=12375\left[\frac{1}{\lambda(\mathring A)}-\frac{1}{\lambda_0(\mathring A)}\right] $
$ =12375\left[\frac{1}{1000}-\frac{1}{2000}\right]=6.2 \mathrm{eV}$
View full question & answer→