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(B) The de-Broglie wavelength of an electron accelerated through a potential $V$ is given by $\lambda_1 = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$.The sho

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$0.1$

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(B) The de-Broglie wavelength of an electron accelerated through a potential $V$ is given by $\lambda_1 = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$.The shortest wavelength of the $X$-rays produced (cut-off wavelength) is given by $\lambda_2 = \frac{hc}{eV}$.The ratio of the wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{h}{\sqrt{2meV}} 	imes \frac{eV}{hc} = \frac{1}{c} \sqrt{\frac{eV}{2m}} = \frac{1}{c} \sqrt{\frac{V}{2} \cdot \frac{e}{m}}$.Substituting the given values: $V = 10^4\, V$,$\frac{e}{m} = 1.8 	imes 10^{11}\, C\, kg^{-1}$,and $c = 3 	imes 10^8\, m/s$.$\frac{\lambda_1}{\lambda_2} = \frac{1}{3 	imes 10^8} \sqrt{\frac{10^4}{2} 	imes 1.8 	imes 10^{11}} = \frac{1}{3 	imes 10^8} \sqrt{0.9 	imes 10^{15}} = \frac{1}{3 	imes 10^8} \sqrt{9 	imes 10^{14}} = \frac{3 	imes 10^7}{3 	imes 10^8} = 0.1$.

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

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