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(D) The wavelength of the $K_\alpha$ line,denoted as $\lambda_{K_\alpha}$,is constant and independent of the operating voltage $V$.The minimum wavelen

Vedclass · Accepted Answer

$\Delta \lambda^{\prime} < 3 \Delta \lambda$

Vedclass · Answer

(D) The wavelength of the $K_\alpha$ line,denoted as $\lambda_{K_\alpha}$,is constant and independent of the operating voltage $V$.The minimum wavelength of the continuous $X$-ray spectrum is given by $\lambda_{min} = \frac{hc}{eV}$.Given $\Delta \lambda = \lambda_{min} - \lambda_{K_\alpha} = \frac{hc}{eV} - \lambda_{K_\alpha}$.When the voltage is changed to $V^{\prime} = V/3$,the new minimum wavelength is $\lambda_{min}^{\prime} = \frac{hc}{e(V/3)} = 3 \frac{hc}{eV} = 3 \lambda_{min}$.The new difference is $\Delta \lambda^{\prime} = \lambda_{min}^{\prime} - \lambda_{K_\alpha} = 3 \lambda_{min} - \lambda_{K_\alpha}$.Since $\lambda_{min} = \Delta \lambda + \lambda_{K_\alpha}$,we substitute this into the expression for $\Delta \lambda^{\prime}$:$\Delta \lambda^{\prime} = 3(\Delta \lambda + \lambda_{K_\alpha}) - \lambda_{K_\alpha} = 3 \Delta \lambda + 3 \lambda_{K_\alpha} - \lambda_{K_\alpha} = 3 \Delta \lambda + 2 \lambda_{K_\alpha}$.Since $\lambda_{K_\alpha} > 0$,it follows that $\Delta \lambda^{\prime} > 3 \Delta \lambda$. Therefore,the correct option is $D$.

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