If 10000, V is applied across an X-ray tube, | vedclass

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(D) 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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$0.3$

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(D) 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 is given by $\lambda_2 = \frac{hc}{eV}$.Taking the ratio of the two wavelengths:$\frac{\lambda_1}{\lambda_2} = \frac{h}{\sqrt{2meV}} 	imes \frac{eV}{hc} = \frac{1}{c} \sqrt{\frac{e^2 V^2}{2meV}} = \frac{1}{c} \sqrt{\frac{eV}{2m}} = \frac{1}{c} \sqrt{\frac{V}{2} \cdot \frac{e}{m}}$.Substituting the given values: $V = 10^4\, V$, $\frac{e}{m} = 1.8 	imes 10^{11}\, C\, kg^{-1}$, and $c = 3 	imes 10^8\, m/s$.$\frac{\lambda_1}{\lambda_2} = \frac{1}{3 	imes 10^8} \sqrt{\frac{10^4}{2} 	imes 1.8 	imes 10^{11}} = \frac{1}{3 	imes 10^8} \sqrt{0.9 	imes 10^{16}} = \frac{0.9486 	imes 10^8}{3 	imes 10^8} \approx 0.316$ (Note: Re-calculating $\sqrt{0.9 	imes 10^{16}} = \sqrt{90 	imes 10^{14}} = 9.48 	imes 10^7$).Actually, $\sqrt{\frac{10^4 	imes 1.8 	imes 10^{11}}{2}} = \sqrt{0.9 	imes 10^{16}} = 0.948 	imes 10^8$. Dividing by $3 	imes 10^8$ gives approximately $0.316$. Given the options, $0.3$ is the closest value.

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 the $X$-rays produced? (Given: $\frac{e}{m}$ for an electron is $1.8 \times 10^{11}\, C\, kg^{-1}$)

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