A potential difference of 20 , kV is applied | vedclass

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(A) The energy of an electron accelerated through a potential difference $V$ is $E = eV$.For the minimum wavelength (maximum energy photon),all kineti

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$0.62$

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(A) The energy of an electron accelerated through a potential difference $V$ is $E = eV$.For the minimum wavelength (maximum energy photon),all kinetic energy of the electron is converted into a single photon: $E = \frac{hc}{\lambda_{\min}}$.Equating the two: $eV = \frac{hc}{\lambda_{\min}}$,which gives $\lambda_{\min} = \frac{hc}{eV}$.Given $V = 20 \, kV = 20 	imes 10^3 \, V$.Using the constant $hc \approx 12400 \, eV \cdot \mathring{A}$,we get:$\lambda_{\min} = \frac{12400 \, eV \cdot \mathring{A}}{20 	imes 10^3 \, eV} = \frac{12400}{20000} \, \mathring{A} = 0.62 \, \mathring{A}$.

$A$ potential difference of $20 \, kV$ is applied across an $X-ray$ tube. The minimum wavelength of $X-rays$ generated (in $\mathring{A}$) is

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