First slide

Matter waves (DC Broglie waves)

Question

Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is

Difficult

A

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

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

$< 2.8 \times 10^{- 10} m$
D

$< 2.8 \times 10^{- 9} m$

Solution

According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted electron is
$K_{\max} = \frac{h c}{λ} - ϕ_{0}$
where $λ$ is the wavelength of incident light and $ϕ_{0}$ is the work function.
$Here, λ = 500 nm, h c = 1240 eVnm$
$and ϕ_{0} = 2.28 eV$
$∴ K_{\max} = \frac{1240 eVnm}{500 nm} - 2.28 eV$
$= 2.48 eV - 2.28 eV = 0.2 eV$
The de Broglie wavelength of the emitted electron is
$λ_{\min} = \frac{h}{\sqrt{2 m K_{\max}}}$
where h is the Planck's constant and m is the mass of the electron.
$As h = 6.6 \times 10^{- 34} J s, m = 9 \times 10^{- 31} kg$
$and K_{\max} = 0.2 eV = 0.2 \times 1.6 \times 10^{- 19} J$
$∴ λ_{\min} = \frac{6.6 \times 10^{- 34} J s}{\sqrt{2 (9 \times 10^{- 31} kg) (0.2 \times 1.6 \times 10^{- 19} J)}}$
$= \frac{6.6}{2.4} \times 10^{- 9} m = 2.8 \times 10^{- 9} m$
$So, λ \geq 2.8 \times 10^{- 9} m$

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