Light of wavelength $500 n m$ is incident on a metal with a work function $2.28 e V$ . The de Borglie wavelength of the emitted electron is:

\leq 2.8 \times 10^{- 12} m

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< 2.8 \times 10^{- 10} m

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< 2.8 \times 10^{- 9} m

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\geq 2.8 \times 10^{- 9} m

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Solution

The correct option is C $\geq 2.8 \times 10^{- 9} m$
Energy of the photon: $E = \frac{12400}{5000} = 2.48 e V$
Work function: $ϕ = 2.28 e V$
$K E_{m a x} = E - ϕ = 2.48 - 2.28 = 0.2 e V$
For electron, $λ_{m i n} = \frac{h}{\sqrt{2 m (K E)_{m a x}}} = 28 A^{0}$
Hence, $λ > 28 A^{0} = 2.8 \times 10^{- 9} m$

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