Verified Solution • 2025–26 Session moderate

QJEEM21Ph0057: JEE Main 2021 Physics - Class 12 JEEM21Ph0057 Solution: When radiation of wavelength is incident on a metallic surface, …

When radiation of wavelength λ\lambda is incident on a metallic surface, the stopping potential of ejected photoelectrons is 4.8 V. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes 1.6 V. The threshold wavelength of the metal is :

Multiple Choice Options

  • A 2 λ\lambda
  • B 4 λ\lambda
  • C 8 λ\lambda
  • D 6 λ\lambda

Step-by-Step Verified Solution

Vs=hv−ϕ{V_s} = hv - \phi

4.8=hcλ−ϕ4.8 = {{hc} \over \lambda } - \phi ..... (i)

1.6=hc2λ−ϕ1.6 = {{hc} \over {2\lambda }} - \phi ..... (ii)

Using above equation (i) - (ii)

3.2=hcλ−hc2λ3.2 = {{hc} \over \lambda } - {{hc} \over {2\lambda }}

3.2=hc2λ3.2 = {{hc} \over {2\lambda }} ..... (iii)

[λ=hc6.4]\left[ {\lambda = {{hc} \over {6.4}}} \right]

Put in equation (ii)

ϕ\phi = 1.6

hcλth=1.6{{hc} \over {{\lambda _{th}}}} = 1.6

λth=hc1.6{\lambda _{th}} = {{hc} \over {1.6}}

=(hc6.4)×4=4λ= \left( {{{hc} \over {6.4}}} \right) \times 4 = 4\lambda

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