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

Question

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 h c = 1230 eV - nm ).

TestHub · Accepted Answer

According to Einstein's photoelectric equation
$k_{m a x} = \frac{h c}{λ} - ϕ$
$\Rightarrow k_{1} = \frac{1230}{800} - ϕ . . . (1)$
And $\Rightarrow k_{2} = \frac{1230}{500} - ϕ$
Since, $k_{2} = 2 k_{1}$
$\Rightarrow 2 k_{1} = \frac{1230}{500} - ϕ . . . (2)$
Eliminating $k_{1}$ from (1) and (2) we get
$0 = \frac{1230}{500} - \frac{1230}{400} + ϕ$
$\Rightarrow ϕ = 0.615 eV$

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