Tag: physics

Questions Related to physics

Momentum of $ \gamma -ray $ photon of energy 3 KeV in kg-m/s will be 

  1. $ 1.6 \times 10^{-19} $

  2. $ 1.6 \times 10^{-2} $

  3. $ 1.6 \times 10^{-24} $

  4. $ 1.6 \times 10^{-27} $


Correct Option: C

What is the momentum of a photon having frequency $1.5 \times 10^{13} Hz $ 

  1. $3.3 \times 10 - 29\, kg\,m/s $

  2. $3.3 \times 10 - 34\, kg\, m/s$

  3. $6.6 \times 10 - 34\, kg\, m/s$

  4. $6.6 \times 10 - 30\, kg\, m/s$


Correct Option: A

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


Correct Option: B
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.

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


Correct Option: C
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.  

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


Correct Option: B
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.

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


Correct Option: A
Explanation:

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

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.


Correct Option: D
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.

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


Correct Option: D
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 $v$.

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


Correct Option: A
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$

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


Correct Option: B
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.