If the minimum wavelength of an X-ray is 2 ma | vedclass

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(B) The minimum wavelength of $X$-rays is given by the Duane-Hunt law: $\lambda_{\min} = \frac{hc}{eV}$.Substituting the values $h = 6.63 	imes 10^{-

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$6.2$

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(B) The minimum wavelength of $X$-rays is given by the Duane-Hunt law: $\lambda_{\min} = \frac{hc}{eV}$.Substituting the values $h = 6.63 	imes 10^{-34} \ J \cdot s$,$c = 3 	imes 10^8 \ m/s$,and $e = 1.6 	imes 10^{-19} \ C$,we get $\lambda_{\min} \approx \frac{12400 \ \mathring{A} \cdot V}{V}$.Given $\lambda_{\min} = 2 \ \mathring{A}$.Therefore,$V = \frac{12400}{\lambda_{\min} (	ext{in } \mathring{A})} \ V$.$V = \frac{12400}{2} \ V = 6200 \ V$.Converting to kilovolts,$V = 6.2 \ kV$.

If the minimum wavelength of an $X$-ray is $2 \mathring{A}$,then the potential difference between the anode and the cathode is ...... $kV$.

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