Question

The de broglie wavelength of an electron (initially at rest) accelerated by an electric potential of 9 volts is given by:
Given: $m_e=9.1\times {10}^{-31}kg,e=1.6\times {10}^{-19}C,$
$h=6.6\times {10}^{-31}Js$

• A. 0.87 nm
• B. 0.62 nm
• C. 0.41 nm
• D. 0.14 nm

Answer 1

The correct option is C 0.41 nm

De broglie relation,
$\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2m(K.E.)}}=\frac{h}{\sqrt{2m\left(qV\right)}}$
where, $\lambda$, $v$, $h$, $q$, $V$ and $m$ are wavelength, velocity, Planck constant, charge, potential and mass of a particle, respectively.

$\lambda =\frac{6.6\times {10}^{-34}}{\sqrt{2\times (9.1\times {10}^{-31})(1.6\times {10}^{-19})\times V}}$

$\lambda =\frac{1.23}{\sqrt{V}}nm$
Given, V = 9 V
$\lambda =\frac{1.23}{\sqrt{9}}nm$ = 0.41 nm