Delta lambda is the difference between the wa | vedclass

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

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 independent of the operating voltage $V$. The minimum wavelength of the continuous $X$-ray spectrum is given by $\lambda_{min} = \frac{hc}{eV}$. The difference is $\Delta \lambda = \lambda_{K_{\alpha}} - \lambda_{min} = \lambda_{K_{\alpha}} - \frac{hc}{eV}$. 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 $\Delta \lambda^{\prime} = \lambda_{K_{\alpha}} - 3 \lambda_{min}$. Since $\lambda_{K_{\alpha}} > \lambda_{min}$,it follows that $\Delta \lambda^{\prime} = \lambda_{K_{\alpha}} - 3 \lambda_{min} Thus,$\Delta \lambda^{\prime} < 3 \Delta \lambda$.

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