Physics

Electrons and Photons

127 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

The work function of aluminum is 4.2eV. Light of wavelength $\displaystyle 2000\mathring {A} $ is incident on it. The maximum energy of emitted electrons will be

  1. Zero

  2. 1.99 volt

  3. 3.2 volt

  4. 6.19 volt
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The work function of aluminium is
$\phi = 4.2 eV = 4.2 \times 1.6 \times 10^{-19}$
$\phi = 6.72 \times 10^{-19} J$
The energy of incident photon is

$h\nu = \dfrac{hc}{\lambda}$
$h\nu = \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{2000\times 10^{-10}}$
$h\nu = 9.9 \times 10^{-19} J$
The maximum energy of photoelectrons is
$E _{max} = h\nu - \phi$
$E _{max} = 9.9 \times 10^{-19} - 6.72 \times 10^{-19}$
$E _{max} = 3.18 \times 10^{-19} J$
$E _{max} = \dfrac{3.18 \times 10^{-19}}{1.6 \times 10^{-19}}$
$E _{max} = 1.99 eV$
So, the answer is option (B).

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

The work function of aluminum is 4.2eV. Light of wavelength $\displaystyle 2000\mathring {A} $ is incident on it. The minimum energy of emitted electrons will be 

  1. Zero

  2. 2.0 eV

  3. 4.2 eV

  4. 6.19 eV
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Energy of the incident light is $6.2 eV$. Now the work function of aluminum is $4.2$. So the  minimum energy of emitted electrons will be $(6.2-4.2)=2$ eV.

So, the answer is option (B).

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

The work function of a metal is X eV. When light of energy 2X is made incident on it then the maximum kinetic energy of emitted photoelectron will be 

  1. $2.5X \ eV$
  2. $2X \ eV$
  3. $X \ eV$
  4. $3X \ eV$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The maximum kinetic energy of emitted photoelectron is given as
$E _{max} = h\nu - \phi$
where,$\nu$ is the frequency of incident light and $\phi$ is photoelectric work function of metal. 
$E _{max} = 2X - X$ ...................(given)
$E _{max} = X \ eV$

So, the answer is option (C).

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

The threshold wavelength of a photosensitive metal is $662.5\ nm$. If this metal is irradiated with a radiation of wavelength $331.3\ nm$, find the maximum kinetic energy of the photoelectrons.

  1. 1.87 eV

  2. 18.7 eV

  3. 17.8 eV

  4. 1.78 eV
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Work function phi = hc/lambda_threshold = (6.63 * 10^-34 * 3 * 10^8) / (662.5 * 10^-9) = 3 * 10^-19 J = 1.875 eV. Incident energy E = hc/lambda = (6.63 * 10^-34 * 3 * 10^8) / (331.3 * 10^-9) = 6 * 10^-19 J = 3.75 eV. K_max = 3.75 - 1.875 = 1.875 eV.

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

The frequency and the intensity of a beam of light falling on the surface of a photoelectric material are increased by a factor of two. This will :

  1. Increase the maximum kinetic energy of the photoelectrons by 2 and photoelectric current by a factor of 1 / 2

  2. Increase the maximum kinetic energy of the photoelectrons, and increase the photoelectric current by a factor of 2

  3. Increase the maximum kinetic energy of the photoelectrons by a factor of 2 and will have no effect on the magnitude of the photoelectric current produced

  4. Not produced any effect on the kinetic energy of the emitted electrons but will increase the photoelectric current by a factor of 2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Doubling the frequency will double the kinetic energy of the photo electron.

Since kinetic energy (E) is given by:
$E=h\nu - \phi$, where $\phi$ is the work function of the material. 
Again the intensity of the light means the increase of number of incident photon on the surface. This will increase the number of ejected photon from the surface in proportionate manner.

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

The frequency of incident light falling on a photosensitive metal plate is doubled, the KE of the emitted photoelectrons is

  1. Double the earlier value

  2. Unchanged

  3. More then doubled

  4. Less than doubled
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

From Einstein's equation for photoelectric effect , the maximum kinetic energy , $K=h\nu-W$

where $\nu=$ frequency of incident light , $h=$ Planck's constant and $W=$ work function of metal. 
When $\nu$ is made double, $K'=2h\nu-W=2(h\nu-W)+W=2K+W$  
Thus, kinetic energy of photoelectrons will be more than doubled when incident frequency is doubled.  

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

When orange light falls on a photo sensitive surface the photocurrent begins to flow.Velocity of emitted electrons will be more when surface is hit by

  1. Red light

  2. Violet light

  3. Thermal radiations

  4. Radio waves
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Frequency of Red light, thermal radiations and radio waves and less than frequency of orange light whereas frequency of violet light is more than frequency of orange light. So, violet light has more energy.

Since energy $=h\nu $
So, when violet light is hit on the surface velocity of emitted electrons will be more.

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

Assertion (A) : For a fixed incident photon energy, photoelectrons have a wide range of energies ranging from zero to the maximum value $K _{max}$
Reason (R) : Initially, the electrons in the metal are at different energy level.

  1. Both A and R are true and R is the correct explanation of A

  2. Both A and R are true but R is not the correct explanation of A

  3. A is true but R is false

  4. A is false but R is true
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Photoelectrons have energy range from zero to the maximum kinetic energy because electrons in the metal have different energy in different energy level.

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

When light falls on a photosensitive surface, electrons are emitted from the surface. The kinetic energy of these electrons does not depend on the:

  1. Wave length of light

  2. Frequency of light

  3. Type of material used for the surface

  4. Intensity of light.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Kinetic energy of photo electrons depend on the wavelength of light, frequency of light, work function of metal, which is property of material, and thickness of the surface layer. Whereas it is independent of intensity of light.

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

The photoelectrons emitted from a metal surface are such that their velocity

  1. Is zero for all

  2. Is same for all

  3. Lies between zero and infinity

  4. Lies between zero and a finite maximum
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Suppose radiation of energy $E + \phi$ is incident on a metal surface
$\phi$ is the work function and is the energy need to eject a photoelectron..
Assume all of E gets converted to Kinetic energy.


$E = \dfrac{1}{2}mv^{2}$ 

Hence, $v = \sqrt{\dfrac{2E}{m}}$

There would be molecules that absorb energy lying from $0$ to $E$.
And hence their velocities would lie between $0$ and $v$.

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

The photoelectric threshold of Tungsten is 2300$ \mathring {A }$. The energy of the electrons ejected from the surface by ultraviolet light of wavelength 18000 $ \mathring {A }$ is 

  1. 0.15 e V

  2. 1.5 e V

  3. 15 e V

  4. 150 e V
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$E _k=d\dfrac{h}{c}\left ( \dfrac{1}{\lambda } -\dfrac{1}{\lambda _o}\right )(in \, e\, V)$

$=\dfrac{6.6\times 10^{34}\times 3\times 10^8}{1.6 \times 10^{-19}}$$\left ( \dfrac{10^{10}}{1800}-\dfrac{10^{10}}{2300} \right )=0.15 e V$

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

When light is incident on a metal surface the maximum kinetic energy of emitted electrons

  1. Vary with intensity of light

  2. Vary with frequency of light

  3. Vary with speed of light

  4. Vary irregulary
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Photoelectric effect
$KE _{max} = hv - W _o$
where,$W _o$ is the work function of the metal
Clearly $KE$ varies linearly with the the frequency of the photons.

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

Ultraviolet radiation of 6.2eV falls on an aluminium surface (work function 4.2eV). The kinetic energy in joule of the faster electron emitted is approximately

  1. $3.2\times 10^{21}$
  2. $3.2\times 10^{-19}$
  3. $3.2\times 10^{-17}$
  4. $3.2\times 10^{-15}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$E _k=E-W _o=6.2-4.2=2.0 e V$
$=2.0 \times 1.6 \times 10^{-19}=3.2\times 10^{-19}\, J$

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

$K _{1} $and $K _{2}$ are the maximum kinetic energies of the photoelectrons emitted when light of wavelength $\lambda _{1} $ and $\lambda _{2} $  respectively are incident on a metallic surface. If $\lambda _{1}= $3$\lambda _{2} $  then

  1. $K _{1}>\dfrac{K _{2}}{3}$
  2. $K _{1}<\dfrac{K _{2}}{3}$
  3. $K _{1}=3K _{2}$
  4. $K _{2}=3K _{1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$K _{1}=\dfrac{hc}{\lambda _{1} }-\phi $-----------(1)

$K _{2}=\dfrac{hc}{\lambda _{2}}-\phi$-----------(2)

$\because\lambda _{1} = 3 \lambda _{2}$

$k _{2}=3\left(\dfrac{hc}{\lambda _{1}}\right)-\phi$---------(3)

$\dfrac{k _{2}}{3}=\dfrac{hc}{\lambda _{1}}-\dfrac{\phi}{3}$----------(4)

$\dfrac{k _{2}}{3}=(k _{1}+\phi)-\dfrac{\phi}{3}$  (by (1))

$\dfrac{k _{2}}{3}=k _{1}+\dfrac{2 \phi}{3}$

So,$k _{1}< \dfrac{k _{2}}{3}$

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

The work function of a metal is $1.6\times 10^{-19}$J. When the metal surface is illuminated by the light of wavelength 6400 $A^{o}$, then the maximum kinetic energy of emitted photoelectrons will be ($h = 6.4 \times 10^{-34} Js$)

  1. $14\times 10^{-19}J$
  2. $2.8\times 10^{-19}J$
  3. $1.4\times 10^{-19}J$
  4. $1.4\times 10^{-19}eV$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$K.E. _{max}=\dfrac{hc}{\lambda }-\phi $


$=\dfrac{6.4\times 10^{-34}\times 3\times 10^{8}}{6.4\times 10^{-7}}-1.6\times 10^{-19}$

$=3\times 10^{-19}-1.6\times 10^{-19}$
$=1.4\times 10^{-19}J.$
So, the answer is option (C).