Physics | VITEEE 2017 | Discussion Page

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

A) $<2.8\times10^{-9}$m

B) $\geq2.8\times10^{-9}$m

C) $\leq2.8\times10^{-12}$m

D) $<2.8\times10^{-10}$m

Option B

Given : work function Φ of metal = 2.28 eV

Wavelength of light λ = 500 nm = 500 x 10^-9m

$KE_{max} = \frac{hc}{\lambda}-\phi$

$KE_{max} = \frac{6.6\times10^{-34}\times3\times10^{8}}{5\times10^{-7}}-2.82$

$KE_{max} =2.48 - 2.28 = 0.2 eV$

$\lambda_{min} =\frac{h}{p}=\frac{h}{\sqrt{2m(KE)_{max}}}$

$\frac{\frac{20}{3}\times10^{-34}}{\sqrt{2\times9\times10^{-31}\times0.2\times1.6\times10^{-19}}}$

$\lambda_{min} =\frac{25}{9}\times10^{-9}$ = 2.80 x 10^-9nm

.'. λ ≥ 2.8 x 10^-9m

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