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If the light of the of wavelength of maximum intensity emitted from surface at temperature $T_{1}$ is used to cause photoelectric emission from a metallic surface, the maximum kinetic energy of the emitted electron is 6 eV, which is 3 times the work function of the metallic surface. If light of wavelength of maximum intensity emitted from a surface at temperature $T_{2,} (T_{2} = 2 T_{1})$ is used, the maximum kinetic energy of the photoelectrons emitted is

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$14 e V$

$4 e V$

$2 e V$

$18 e V .$

answer is A.

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Detailed Solution

$\begin{array}{l} ϕ = 2 e V \\ \frac{h c}{λ_{1}} = 8 e V \\ T_{2} = 2 T_{1} \end{array}$

If $λ_{1}$ is the wavelength corresponding to maximum intensity at

$T_{1} & λ_{2} a t T_{2} :$

Then $\begin{array}{l} λ_{2} = \frac{λ_{1}}{2} (b y w e i n' s d i s p l a c e m e n t L a w) \\ \frac{h c}{λ_{2}} = \frac{h c}{λ_{1}} = 16 e V \\ ϕ = 2 e V \\ ∴ K . E_{\max} = \frac{h c}{λ_{1}} - ϕ = 14 e V . \end{array}$

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