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

  • Option 1)

    +3 V

  • Option 2)

    +4 V

  • Option 3)

    -1 V

  • Option 4)

    -3 V

 

Answers (1)
V Vakul

As we learnt in 

Conservation of energy -

h\nu = \phi _{0}+\frac{1}{2}mv^{2}_{max}

h\nu = h\nu _{0}+\frac{1}{2}mv^{2}_{max}

 

h\left ( \nu-\nu _{0} \right )=\frac{1}{2}mv^{2}_{max}

where, h - Plank's constant\ \nu - Frequency\ \nu_{0} - threshold\ frequency\ \phi_{0}- work function

- wherein

 

 ev_{s}=hv-\phi =K.E_{max}

or\: 2ev=5ev-\phi

\therefore \phi =3ev

When electron is given 6ev by photon

its maximum kinetic energy will be  K.Emax = 6ev -3ev = 3ev

=> The stopping potential must be the minimum potential to stop this electron from reaching anode A. Then stopping potential of A relative

to C = -3ev


Option 1)

+3 V

Option is incorrect

Option 2)

+4 V

Option is incorrect

Option 3)

-1 V

Option is incorrect

Option 4)

-3 V

Option is correct

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