Verified Solution • 2025–26 Session moderate

QJEEM21Ph4d68: JEE Main 2021 Physics - Class 12 JEEM21Ph4d68 Solution: A monochromatic neon lamp with wavelength of 670.5 nm…

A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which has a stopping voltage of 0.48 V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6 nm?

Multiple Choice Options

  • A 0.96 V
  • B 1.25 V
  • C 0.24 V
  • D 1.5 V

Step-by-Step Verified Solution

kEmax⁡=hcλi+ϕk{E_{\max }} = {{hc} \over {{\lambda _i}}} + \phi

or eVo=hcλi+ϕe{V_o} = {{hc} \over {{\lambda _i}}} + \phi

when λ\lambda i = 670.5 nm ; Vo = 0.48

when λ\lambda i = 474.6 nm ; Vo = ?

So,

e(0.48)=1240670.5+ϕe(0.48) = {{1240} \over {670.5}} + \phi ..... (1)

e(Vo)=1240474.6+ϕe({V_o}) = {{1240} \over {474.6}} + \phi .....(2)

(2) −- (1)

e(Vo−0.48)=1240(1474.6−1670.5)eVe({V_o} - 0.48) = 1240\left( {{1 \over {474.6}} - {1 \over {670.5}}} \right)eV

Vo=0.48+1240(670.5−474.6474.6×670.5){V_o} = 0.48 + 1240\left( {{{670.5 - 474.6} \over {474.6 \times 670.5}}} \right) Volts

Vo = 0.48 + 0.76

Vo = 1.24 V ≃\simeq 1.25 V

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