What is the de Broglie wavelength of an electron of energy 180 eV? Mass of electron $= 9 \times 10^{- 31}$ kg and Planck's constant $= 6.6 \times 10^{- 34}$ Js.

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$1.3 Å$

$1.8 Å$

$0.9 Å$

$0.5 Å$

answer is B.

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Detailed Solution

$K = 180 eV = 180 \times 1 .6 \times 10^{- 19} J = 2 .88 \times 10^{- 17} J$

Now $λ = \frac{h}{\sqrt{2 mK}}$

$\begin{matrix} ∴ λ = \frac{6 .6 \times 10^{- 34}}{{[2 \times 9 \times 10^{- 31} \times 2 .88 \times 10^{- 17}]}^{1 / 2}} \\ = \frac{6 .6 \times 10^{- 34}}{7 .2 \times 10^{- 24}} = 0 .9 \times 10^{- 10} m \end{matrix}$

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