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(B) The minimum wavelength $\lambda_{min}$ of $X$-rays produced by an accelerating voltage $V$ is given by the formula: $\lambda_{min} = \frac{hc}{eV}

Vedclass · Accepted Answer

$> 124.2 \ kV$

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(B) The minimum wavelength $\lambda_{min}$ of $X$-rays produced by an accelerating voltage $V$ is given by the formula: $\lambda_{min} = \frac{hc}{eV} = \frac{12400 \ 	ext{eV} \cdot \mathring{A}}{eV} \approx \frac{12400}{V(	ext{volts})} 	imes 10^{-10} \ m$.Given $\lambda_{min} = 10^{-11} \ m = 10^{-10} \ \mathring{A}$.Substituting the values: $10^{-11} = \frac{12400 	imes 10^{-10}}{V}$.$V = \frac{12400 	imes 10^{-10}}{10^{-11}} = 124000 \ V = 124 \ kV$.Since the question specifies that the minimum wavelength should be $10^{-11} \ m$,the energy of the electrons must be at least $124 \ kV$. Therefore,the accelerating voltage must be at least $124.2 \ kV$ (or greater) to produce $X$-rays of this wavelength or shorter.

For the study of bone fractures,the minimum wavelength of $X$-rays should be $10^{-11} \ m$. The accelerating voltage for electrons in the $X$-ray machine will be .......

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