A potential difference of $10^{4}$ V is applied across an $X$-ray tube. The ratio of the de-Broglie wavelength of $X$-rays produced is $(\frac{e}{m}\ for\ electron = 1.8 imes 10^{11}Ckg^{- 1})$

Question

A potential difference of $10^{4}$ V is applied across an $X$-ray tube. The ratio of the de-Broglie wavelength of $X$-rays produced is $(\frac{e}{m}\ for\ electron = 1.8 	imes 10^{11}Ckg^{- 1})$

Dr. Abubacker · Accepted Answer

(b) de-Broglie wavelength, $\lambda = \frac{h}{\sqrt{2meV}}$
X-ray wavelength, $\lambda_{2} = \frac{hc}{eV}$
$	herefore\ \frac{\lambda_{1}}{\lambda_{2}} = \frac{eV}{c\sqrt{2meV}} = \frac{1}{c}\sqrt{\frac{1}{2}\left( \frac{e}{v} ight)V}$
$= \frac{1}{3 	imes 10^{8}}\sqrt{\frac{1}{2} 	imes 1.8 	imes 10^{11} 	imes 10^{4}} = \frac{1}{10}$

Q. A potential difference of $10^{4}$ V is applied across an $X$-ray tube. The ratio of the de-Broglie wavelength of $X$-rays produced is $(\frac{e}{m}\ for\ electron = 1.8 \times 10^{11}Ckg^{- 1})$

Explanation

Reviewed by Dr. Abubacker

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