First slide

Wave nature of matter

Question

If 10000 V is applied across an X-ray tube, what will be the ratio of de Broglie wave length of the incident electrons to the shortest wavelength of X-ray produced ( $\frac{e}{m}$ for electron is $1.8 \times 10^{11} C {kg}^{- 1}$ )

Moderate

A

1
B

0.1
C

0.2
D

0.3

Solution

For the incident electron, $\frac{1}{2} m v^{2} = e V or P = \sqrt{2 m K} = \sqrt{2 m e V}$
$\begin{array}{l} ∴ de Broglie wavelength, λ_{1} = \frac{h}{p} = \frac{h}{\sqrt{2 m e V}} \\ Shortest X -ray wavelength λ_{2} = \frac{h c}{e V} \\ ∴ \frac{λ_{1}}{λ_{2}} = \frac{1}{c} \sqrt{(\frac{V}{2}) (\frac{e}{m})} = \frac{\sqrt{\frac{10^{4}}{2} \times 1.8 \times 10^{11}}}{3 \times 10^{8}} = 0.1 \end{array}$

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