First slide

Wave nature of matter

Question

What is the de Broglie wavelength of the wave associated with an electron that has been accelerated through a potential difference of 50.0 V?

Moderate

A

$2.7 \times 10^{- 10}$
B

$1.74 \times 10^{- 10}$
C

$3.6 \times 10^{- 9}$
D

$4.9 \times 10^{- 11}$

Solution

The gain of kinetic energy by an electron is eV.
$\begin{array}{l} \frac{1}{2} m v^{2} = e V \\ v = \sqrt{\frac{2 e V}{m}} = \sqrt{\frac{2 (1.60 \times 10^{- 19}) (50)}{(9.11 \times 10^{- 31})}} \\ = 4.19 \times 10^{6} {ms}^{- 1} \end{array}$
Thus, the electron's de Broglie wavelength is
$\begin{matrix} λ = \frac{h}{m v} = \frac{6.63 \times 10^{- 34}}{(9.1 \times 10^{- 31}) (4.19 \times 19^{6})} \\ = 1.74 \times 10^{- 10} m \end{matrix}$

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