Find the maximum magnitude of the linear momentum of a photoelectron emitted when a wavelength of 400 nm falls on a metal with work function 2.5 eV.

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Solution

Given:
Wavelength of light, $λ$ = 400 nm = 400 $\times$ 10^{$- 9$} m
Work function of metal, $ϕ$ = 2.5 eV
From Einstein's photoelectric equation,
$Kinetic energy = \frac{h c}{λ} - ϕ$
Here, c = speed of light
h = Planck's constant
$∴ K . E . = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{4 \times 10^{- 7} \times 1.6 \times 10^{- 19}} - 2.5 eV = 0.605 eV$
Also, K.E. = $\frac{p^{2}}{2 m}$ ,
where p is momentum and m is the mass of an electron.
$∴ p^{2} = 2 m \times K . E . \Rightarrow p^{2} = 2 \times 9.1 \times 10^{- 31} \times 0.605 \times 1.6 \times 10^{- 19} \Rightarrow p = 4.197 \times 10^{- 25} kg - m / s$

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