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When light of wavelength 248 nm falls on a metal ....

When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electron is _ _ _ _ $A^\circ $. (Round off to the nearest integer)

[Use $\sqrt 3 = 1.73$, $h=6.63 \times 10^{-34} Js$, $m_e = 9.1 \times 10^{-31} Kg $, $c = 3.0 \times 10^8 ms^{-1}$, $1 eV = 1.6 \times 10^{-19} J$]

Solution

We have, $h\nu = \phi + KE $

$\therefore KE =  \frac {hc}{\lambda} - \phi $

$\Rightarrow KE = \frac {6.63\times 10^{-34} \times 3.0 \times 10^8 }{248 \times 10^{-9}} - 3.0 \times 1.6 \times 10^{-19} $

$\Rightarrow KE = 8 \times 10^{-19} - 4.8 \times 10^{-19} = 3.2 \times 10^{-19} J$

de-Broglie wavelength $\lambda = \frac {h}{\sqrt {2.KE.m_e }} = \frac {6.63 \times 10^{-34}}{\sqrt {2 \times 3.2 \times 10^{-19} \times 9.1 \times 10^{-31}}}$

$\Rightarrow \lambda = \frac {6.63 \times 10^{-34}}{7.63 \times 10^{-25}} = 8.7\times 10^{-10} m = 8.7 A ^\circ \approx 9 A ^\circ $

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