Physics

Electrons and Photons

158 Questions

Electrons and photons are fundamental concepts in modern physics, focusing on quantum interactions and the photoelectric effect. The topic involves calculating work functions, kinetic energy, and photon wavelengths. These physics questions are highly relevant for competitive test preparation.

Photoelectric effectWork function calculationsPhoton energy and wavelengthElectron emissionAtomic transitions

Electrons and Photons Questions

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Light of wavelength $\lambda$ strikes a photo sensitive surface and electrons are ejected with kinetic energy $E$. If the kinetic energy is to be increased to $2E$, then the wavelength must be changed to $\lambda'$, where :

  1. $\lambda' > \lambda$
  2. $\lambda' = \dfrac {\lambda}{2}$
  3. $\lambda' = 2\lambda$
  4. $\dfrac {\lambda}{2} > \lambda' > \lambda$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have, $E _{k} = \dfrac {hc}{\lambda} - \phi _{0}$


and $2E _{k} = \dfrac {hc}{\lambda'} - \phi _{0}$

By the two relations, we have

$\lambda' > \dfrac {\lambda}{2}$

and $\lambda' < \lambda$

So, $\dfrac {\lambda}{2} < \lambda' < \lambda$.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

A photo-cell is illuminated by a source of light, which is placed at a distance $d$ from the cell. If the distance become $d/2$, then number of electrons emitted per second will be : 

  1. Remain same

  2. Four times

  3. Two times

  4. One-fourth
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Intensity, $I=\dfrac { E }{ At } \ $


where E is the total energy of source,


A is area of illuminated surface,

t is time

$\therefore I=\dfrac { E }{ 4\pi { { { r }^{ 2 } }t } } \\ \dfrac { { I } _{ 1 } }{ { I } _{ 2 } } =\dfrac { E }{ 4\pi { { { r } _{ 1 }^{ 2 } }t } } \times \dfrac { 4\pi { { { r } _{ 2 }^{ 2 } }t } }{ E } \\ =\dfrac { { r } _{ 2 }^{ 2 } }{ { r } _{ 1 }^{ 2 } } \\ \dfrac { { I } _{ 1 } }{ { I } _{ 2 } } =\dfrac { 4 }{ 1 } \\ \Rightarrow \dfrac { { I } _{ 2 } }{ { I } _{ 1 } } =\dfrac { 1 }{ 4 } $

Since, $\propto$ number of photoelectrons emitted

Therefore, Number of electrons emitted is a quarter of the initial number.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

A monochromatic source of light is placed at a large distance $d$ from a metal surface. Photoelectrons are ejected at rate $n$, the kinetic energy being $E$. If the source is brought nearer to distance $d/2$, the rate and kinetic energy per photoelectron become nearly :

  1. 2n and 2E

  2. 4n and 4E

  3. 4n and E

  4. n and 4E
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The rate is inversely proportional to square of the distance: $n\propto \dfrac{1}{{r}^{2}}$


So the new rate will be $4n$

Kinetic energy is not related to the distance, hence it will remain same $E$.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

The work function of a certain metal is $3.31 \, \times \,  10^{-19} J.$ Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength 5000 A is
$(Given\, h  = 6.62\times10^{-34} J-s, \, c= 3\times10^{-8} \, ms^{-1},\, e=  1.6 \times10^{-19} \, C)$

  1. $248 eV$
  2. $0.41 eV$
  3. $2.07 eV$
  4. $0.82 eV$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Work function $W _0 \, = \, 3.31 \, \times \, 10^{-19} \, J$
Wavelength of incident radiation
 $\lambda \, = \, 5000 \, \times \, 10^{-10} \, m$
$E \, = \, W _0 \, + \, KE$
(According to Einstein equation)
$\displaystyle \frac{hc}{\lambda} \, 3.31 \, \times \, 10^{-19} \, + \, KE$
$KE \, = \, - \, 3.31 \, \times \, 10^{-19} + \, \displaystyle \frac{6.62 \, \times \, 10^{-34} \, \times \, 3 \, \times \, 10^8}{5000 \, \times \, 10^{10}}$
$= \, - \, 3.31 \, \times \, 10^{-19} \, + \, 10^{-19} \, + \, \displaystyle \frac{6.62 \, \times \, 3}{5} \, \times \, 10^{-19}$
$= \, (- \, 3.31 \, \times \, 1.324 \, \times \, 3) \, \times \, 10^{-19}$
$= \, (3.972 \, - \, 3.31) \, \times \, 10^{-19} \, = \, 0.662 \, \times \, 10^{-19} \, J$
$\Rightarrow \, E \, = \, \displaystyle \frac{0.662 \, \times \, 10^{19}}{1.6 \, \times \, 10^{-19}} \, = \, 0.41 eV$

Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

The phenomenon of pair production is :

  1. The production of an electron and a positron from $\gamma$ radiation

  2. Ejection of an electron from a metal surface when exposed to ultraviolet light

  3. Ejection of an electron from a nucleus

  4. Ionization of a neutral atom
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Pair production :
Gamma rays of sufficient energy when passing near a nucleus disappear and materialize into pair of an electron and a positron.
To have pair production, minimum energy of $\gamma$ - ray radiation is 1.02MeV.

Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

A photon of $1.7 \times 10 ^{-13}$ joule is absorbed by a material under special circumstances. The correct statement is :

  1. Electron of the atoms of absorbed material will go the higher energy states.

  2. Electron and positron pair will be created

  3. Only positron pair will be produced

  4. Photoelectric effect will occur and electron will be produced
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For electron and positron pair production, minimum energy is $1.02\ MeV$.
Energy of photon is given: $ 1.7 \times 10^{-3} J=\dfrac{1.7 \times 10^{-13}}{1.6 \times 10^{-19}}$$=1.06 \, MeV$.
Since energy of photon is greater than 1.02 MeV, electron positron pair will be created.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

When a certain photosensistive surface is illuminated with monochromatic light of frequency v, the stopping potential for the photo current is-${ V } _{ 0 }/2.$ When the surface frequency v/2, the stopping potential is -${ V } _{ 0 }.$ The threshold frequency for photoelectric emission is

  1. $\dfrac { 5v }{ 3 } $
  2. $2v$
  3. $\dfrac { 4 }{ 3 } v$
  4. $\dfrac { 3v }{ 2 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

A metal plate is placed 2 m away from a monochromatic light source of 1 mW power. Assuming that an electron in metal collects its energy from a circular area of the plate as large as 10 atomic diameters (${10^{ - 9}}\;m$) in radius, calculate how long it will take for such a 'target' to 'soak off' % eV of energy for its emission from the metal?

  1. 1.5 hr

  2. 2.5 hr

  3. 3.5 hr

  4. 4.5 hr
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The power is spread over a sphere of area 4π(2)^2 = 16π m^2 at distance 2m. The electron collects from area π(10^-9)^2 m^2, so the fraction is 10^-18/16 = 6.25×10^-20. Power on target = 1 mW × 6.25×10^-20 = 6.25×10^-23 W. For 5 eV (5 × 1.6×10^-19 J), time = 8×10^-19 / 6.25×10^-23 ≈ 12800 s ≈ 3.5 hr. The question likely meant 5 eV (typo: % eV).

Multiple choice botany cell and cellular organization cell theory microscope cell and its discovery

A major breakthrough in the studies of cells came with the development of electron microscope. This is because

  1. Electron microscope is more powerful than the light microscope as it uses a beam of electrons which has wavelength much longer than that of photons.

  2. Resolution power of electron microscope is much higher than that of light microscope.
  3. Resolution power of electron microscope is $200-350$nm as compared to $0.1-02$nm for the light microscope.

  4. Electron beam can pass through thick materials whereas light microscopy requires thin sections.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The electron microscope provides much higher resolution than a light microscope because electrons have a much shorter wavelength than visible light photons, allowing for the visualization of much smaller structures.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

If accelerating potential increases from $20\ KV$ to $80\ KV$ in an electron microscope, its resolving power $R$ would change to

  1. $\dfrac{R}{4}$
  2. $4R$
  3. $2R$
  4. $\dfrac{R}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\dfrac{1}{2}mv^{2}= eV$

$mv= \sqrt{2eVm}$

And $\lambda = \dfrac{h}{mV}$

$\dfrac{\lambda _{0}}{\lambda _{1}}= \dfrac{\sqrt{2eV _{1}m}}{\sqrt{eV _{2}m}}$

$\dfrac{\lambda _{2}}{\lambda _{1}}= \dfrac{1}{2}$

$\therefore \lambda _{2}=\dfrac{\lambda _{1}}{2}$

$R\ \propto \dfrac{1}{\lambda}$

so $R$ would change to $2R$.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

In an electron microscope the accelerating voltage is increased from 20 kV to 80 kV, the resolving power of the microscope will change from R to

  1. $2 R$
  2. $\dfrac{R}{2}$
  3. $4R$
  4. $3R$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Electron microscope is a microscope that can magnify very small details with high resolving power due to the use of electrons as the source of illumination. Since the wavelength of electrons are 100,000 times shorter than visible light the electron microscopes have greater resolving power
We have the Abbe's formula as resolution limit $d=\dfrac{0.61\lambda}{NA}$(NA is the numerical aperture)


The resolving power increases when d, the minimum distance that can be seen between two points in the image, decreases. Thus, according to the formula the resolving power is inversely proportional to the wavelength.

Resolving Power  $\propto \dfrac{1}{\lambda}$

A higher voltage will give the electrons a higher speed. Thus the electrons will have a smaller de Broglie wavelength according to the equation,  $\lambda=h/mv$

$\lambda\propto\dfrac{1}{\sqrt V}$

Thus we get Resolving power $\propto \sqrt{V}$

$ \implies\dfrac{R}{R'} = \sqrt{\dfrac{20}{80}} $

Thus, $R' = 2R$

Multiple choice physics static electricity properties of charges charge properties of charge

Electron accelerated by potential $V$ are diffracted from a crystal. If $d=1 A$ and $i = 30^\circ $. $V$ should be about  $h = 6.6 \times {10^{ - 24}}Js\,{m _e} = 9.1 \times {10^{ - 33}}kg.e = 1.6 \times {10^{ - 19}}C$

  1. $2000 V$
  2. $50 V$
  3. $500 V$
  4. $1000 V$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$d=1A$

$i={ 30 }^{ o  }$
$\theta ={ 60 }^{ o  }$
$h=6.6\times { 10 }^{ -34 }Js$
${ m } _{ e }=9.1\times { 10 }^{ -31 }kg$
$e=1.6\times { 10 }^{ -19 }c$
$n\lambda =2dsin\theta $
$\lambda =\dfrac { 2\times 1A\times sin{ 60 }^{ o  } }{ 1 } $
$\lambda =\sqrt { 3A } $
$\sqrt { V } =\dfrac { 1.27\times { 10 }^{ -10 } }{ \sqrt { 3 } \times { 10 }^{ -10 } } =50.18volts$