Verified Solution • 2025–26 Session moderate

QJEEM16Pha98a: JEE Main 2016 Physics - Class 12 JEEM16Pha98a Solution: A photoelectric surface is illuminated successively by…

A photoelectric surface is illuminated successively by monochromatic light of wavelengths λ\lambda and λ2.{\lambda \over 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 is :

Multiple Choice Options

  • A hc3λ{{hc} \over {3\lambda }}
  • B hc2λ{{hc} \over {2\lambda }}
  • C hcλ{{hc} \over {\lambda }}
  • D 3 hcλ{3\,{hc} \over {\lambda }}

Step-by-Step Verified Solution

We know,

Einstein's photo electric equation,

(KE)max = hcλ{{hc} \over \lambda } −- ϕ\phi 0

In first case,

K = hcλ{{hc} \over \lambda } −- ϕ\phi 0           . . .(1)

In second case,

3K = 2hcλ{{2hc} \over \lambda } −- ϕ\phi 0           . . .(2)

⇒\Rightarrow     3(hcλ−ϕ0)3\left( {{{hc} \over \lambda } - {\phi _0}} \right) = 2hcλ−ϕ0{{2hc} \over \lambda } - {\phi _0}

⇒\Rightarrow     3hcλ{{3hc} \over \lambda } −- 3 ϕ\phi 0 = 2hcλ{{2hc} \over \lambda } −- ϕ\phi 0

⇒\Rightarrow     hcλ{{hc} \over \lambda } = 2 ϕ\phi 0

⇒\Rightarrow     ϕ\phi   =   hc2λ{{hc} \over {2\lambda }}

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