The de Broglie wavelength of an electron is $0.4 \times 10^{- 10} m$ when its kinetic energy is $1.0 k e V$ . Its wavelength will be $1.0 \times 10^{- 10} m$ , when its kinetic energy is

0.2 k e V

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0.8 k e V

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0.63 k e V

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0.16 k e V

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Solution

The correct option is D $0.16 k e V$
$λ = \frac{h}{p}, λ = \frac{h}{\sqrt{2 m k}}$
so, $λ \propto \frac{1}{\sqrt{k}}$
$\Rightarrow \frac{0.4 \times 10^{- 10}}{1.0 \times 10^{- 10}} = \frac{\sqrt{k}}{\sqrt{1}} \Rightarrow k = 0.16 k e V$

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