A metal exposed to light of wavelength 800 nm and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500 nm is used. The work function of the metal is (Take hc = 1230 eV-nm).

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0.615 eV

1.23 eV

1.537 eV

2.46 eV

answer is C.

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Detailed Solution

$\begin{array}{l} k_{1} = \frac{1230}{800} - ϕ . . . (1) \\ k_{2} = 2 k_{1} = \frac{1230}{500} - ϕ . . . (2) \end{array}$

Eliminating k₁ from (1) and (2) we get

$\begin{array}{l} 0 = \frac{1230}{500} - \frac{1230}{400} + ϕ \\ ϕ = 0 .615 eV \end{array}$

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