First slide

X-rays

Question

The potential difference applied to an X-ray tube is V. The ratio of the de Broglie wavelength of electron to the minimum wavelength of X-ray is directly proportional to

Easy

A

V
B

$\sqrt{V}$
C

$V^{3 / 2}$
D

$V^{7 / 2}$

Solution

$\begin{array}{l} d e B r o g l i e w a v l e n g t h = λ = \frac{h}{\sqrt{2 m e V}} \\ L e t s a y c u t o f f w a v e l e n g t h = λ_{min} \\ \Rightarrow \frac{h c}{λ_{min}} = eV \\ M u l t i p l y i n g w e g e t, \\ λ \times \frac{h c}{λ_{min}} = \frac{h}{\sqrt{2 m e V}} eV \\ \Rightarrow \frac{λ}{λ_{min}} \propto \sqrt{V} \end{array}$

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