An electromagnetic wave of wavelength $λ$ is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de Brogue wavelength $λ^{'}$ , then

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Answer 1

Kinetic energy of emitted electron = Energy of incident photon
i.e. $\frac{1}{2} m v^{2} = h ν$

or $\frac{p ^{2}}{2 m} = \frac{h c}{λ}$

$[∵ m v = p, ν = \frac{c}{λ}]$
or $p = \frac{2 m h c}{λ}$

de-Broglie wavelength of emitted electrons

$λ^{'} = \frac{h}{p} = \frac{h}{\frac{2 m h c}{λ}}$ or

$λ^{'} = \frac{h λ}{2 m c}$ $∴ λ = \frac{2 m c}{h} λ^{' 2}$

Answer 2

Kinetic energy of emitted electron = Energy of incident photon
i.e.

\frac{1}{2} m v^{2} = h ν

or

\frac{p ^{2}}{2 m} = \frac{h c}{λ}

[∵ m v = p, ν = \frac{c}{λ}]

or

p = \frac{2 m h c}{λ}

de-Broglie wavelength of emitted electrons

λ^{'} = \frac{h}{p} = \frac{h}{\frac{2 m h c}{λ}}

or

λ^{'} = \frac{h λ}{2 m c}

∴ λ = \frac{2 m c}{h} λ^{' 2}