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

Question

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

Dr. Abubacker · Accepted Answer

(b) For incident electron $\frac{1}{2}mv^{2} = eV$ or$\ p^{2} = 2meV$
$	herefore$ de-Broglie wavelength $\lambda_{1} = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$
Shortest $X$-ray wavelength $\lambda_{2} = \frac{hc}{eV}$
$	herefore\frac{\lambda_{1}}{\lambda_{2}} = \frac{1}{c}\sqrt{\left( \frac{V}{2} ight)\left( \frac{e}{m} ight)} = \frac{\sqrt{\frac{10^{4}}{2} 	imes 1.8 	imes 10^{11}}}{3 	imes 10^{8}} = 0.1$

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

Explanation

Reviewed by Dr. Abubacker

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