Q 1 :    

If ϕ is the work function of photosensitive material in eV and light of wavelength of numerical value λ=hce metre, is incident on it with energy above its threshold value at an instant, then the maximum kinetic energy of the photo-electron ejected by it at that instant (Take h - Planck’s constant, c - velocity of light in free space) is (in SI units)   [2024]
 

  • e+2ϕ

     

  • 2e-ϕ

     

  • e-ϕ

     

  • e+ϕ

     

(3)

According to Einstein photoelectric equation,

Photon energy=K.E.+Work function

Here, photon energy =e eV

Work function =ϕ eV

   e=K.E.+ϕ    K.E.=e-ϕ



Q 2 :    

The maximum kinetic energy of the emitted photoelectrons in photoelectric effect is independent of:               [2023]

  • work function of material

     

  • intensity of incident radiation

     

  • frequency of incident radiation

     

  • wavelength of incident radiation

     

(2)

Using photoelectric equation,  

E=hυ-ϕ0

Maximum kinetic energy of emitted electron is independent of intensity of radiation.  

The intensity of incident radiation varies with photo current if threshold frequency requirement is fulfilled.



Q 3 :    

When two monochromatic light of frequency, υ and υ2 are incident on a photoelectric metal, their stopping potential becomes Vs2 and Vs respectively. The threshold frequency for this metal is              [2022]
 

  • 2υ

     

  • 3υ

     

  • 23υ

     

  • 32υ

     

(4)

Let the threshold frequency be υ0.

By using the equation of photoelectric effect, E=hυ0+eV0

Case I : hυ=hυ0+eVs2                                               ...(i)

Case II: hυ2=hυ0+eVs                                                ...(ii)

hυ2=-eVs2

-hυ=eVs  (put in (i))

hυ=hυ0-hυ2    υ0=32υ



Q 4 :    

The work function of a photosensitive material is 4.0 eV. The longest wavelength of light that can cause photon emission from the substance is (approximately)         [2019]

  • 3100 nm

     

  • 966 nm

     

  • 31 nm

     

  • 310 nm

     

(4)

Required wavelength of light,  

λ0=hcϕ=1240 eV nm4 eV310 nm



Q 5 :    

When the light of frequency 2υ0 (where υ0 is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is v1. When the frequency of the incident radiation is increased to 5υ0, the maximum velocity of electrons emitted from the same plate is v2. The ratio of v1 to v2 is              [2018]
 

  • 1 : 2

     

  • 1 : 4

     

  • 4 : 1

     

  • 2 : 1

     

(1)

According to the Einstein’s photoelectric equation,

E=W0+12mv2

When frequency of incident light is 2υ0.

h(2υ0)=hυ0+12mv12    hυ0=12mv12                  ...(i)

When frequency of incident light is 5υ0

h(5υ0)=hυ0+12mv22    4hυ0=12mv22                ...(ii)

Dividing (i) by (ii),  14=v12v22  or  v1v2=12

 



Q 6 :    

Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is              [2016]

  • +3 V

     

  • +4 V

     

  • –1 V

     

  • –3 V

     

(4)

According to Einstein's photoelectric equation, maximum kinetic energy of photoelectrons,

         KEmax=Eν-ϕ

or      2=5-ϕ    ϕ=3eV

When  Eν=6 eV then, KEmax=6-3=3 eV

or   e(Vcathode-Vanode)=3 eV

or   Vcathode-Vanode=3V=-Vstopping

   Vstopping=-3 V



Q 7 :    

When a metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is V4. The threshold wavelength for the metallic surface is                            [2016]
 

  • 52λ

     

  • 3λ

     

  • 4λ

     

  • 5λ

     

(2)

According to Einstein's photoelectric equation,

        eVs=hcλ-hcλ0

   As per question, eV=hcλ-hcλ0                    ...(i)

        eV4=hc2λ-hcλ0                                                    ...(ii)

From equations (i) and (ii), we get

         hc2λ-hc4λ=hcλ0-hc4λ0hc4λ=3hc4λ0     or   λ0=3λ



Q 8 :    

A photoelectric surface is illuminated successively by monochromatic light of wavelength λ and λ/2. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is (h = Planck’s constant, c = speed of light)           [2015]
 

  • 2hcλ

     

  • hc3λ

     

  • hc2λ

     

  • hcλ

     

(3)

Let ϕ0 be the work function of the surface of the material.
According to Einstein’s photoelectric equation, the maximum kinetic energy of the emitted photoelectrons in the first case is

            Kmax1=hcλ-ϕ0

and that in the second case is

           Kmax2=hcλ2-ϕ0=2hcλ-ϕ0

But  Kmax2=3Kmax1             (given)

    2hcλ-ϕ0=3(hcλ-ϕ0);  2hcλ-ϕ0=3hcλ-3ϕ0

         3ϕ0-ϕ0=3hcλ-2hcλ  or  2ϕ0=hcλ  or  ϕ0=hc2λ



Q 9 :    

A certain metallic surface is illuminated with monochromatic light of wavelength, λ. The stopping potential for photoelectric current for this light is 3V0. If the same surface is illuminated with light of wavelength 2λ, the stopping potential is V0. The threshold wavelength for this surface for photoelectric effect is            [2015]
 

  • λ4

     

  • λ6

     

  • 6λ

     

  • 4λ

     

(4)

 

 



Q 10 :    

When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is          [2014]

  • 0.65 eV

     

  • 1.0 eV

     

  • 1.3 eV

     

  • 1.5 eV

     

(2)

According to Einstein's photoelectric equation, the kinetic energy of emitted photoelectrons is

      K=hυ-ϕ0

where hυ is the energy of incident radiation and ϕ0 is the work function of the metal.

As per question,

           0.5eV=hυ-ϕ0                                ...(i)

            0.8eV=1.2hυ-ϕ0                         ...(ii)

On solving eqns. (i) and (ii), we get

            ϕ0=1.0 eV