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(B) The minimum wavelength of an $X$-ray spectrum is given by $\lambda_{\min} = \frac{hc}{eV}$.Let the initial voltage be $V_1 = V$ and the final volt

Vedclass · Accepted Answer

$\approx 16 \, kV$

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(B) The minimum wavelength of an $X$-ray spectrum is given by $\lambda_{\min} = \frac{hc}{eV}$.Let the initial voltage be $V_1 = V$ and the final voltage be $V_2 = 1.5V$.The initial wavelength is $\lambda_1 = \frac{hc}{eV}$ and the final wavelength is $\lambda_2 = \frac{hc}{e(1.5V)} = \frac{hc}{1.5eV}$.Since the wavelength decreases as voltage increases, the shift is $\Delta \lambda = \lambda_1 - \lambda_2 = 26 \, pm$.$\Delta \lambda = \frac{hc}{eV} - \frac{hc}{1.5eV} = \frac{hc}{eV} (1 - \frac{1}{1.5}) = \frac{hc}{eV} (1 - \frac{2}{3}) = \frac{hc}{3eV} = 26 \, pm$.Using $hc \approx 12400 \, eV \cdot$ $\mathring{A}$ $= 1240 \, eV \cdot nm = 1.24 	imes 10^{-6} \, eV \cdot m$ or $1240000 \, eV \cdot pm$:$V = \frac{hc}{3e \cdot \Delta \lambda} = \frac{1240000 \, eV \cdot pm}{3 \cdot 26 \, pm} \approx \frac{1240000}{78} \approx 15897 \, V \approx 16 \, kV$.

If a voltage applied to an $X$-ray tube is increased to $1.5$ times, the minimum wavelength $(\lambda_{\min})$ of an $X$-ray continuous spectrum shifts by $\Delta \lambda = 26 \, pm$. The initial voltage applied to the tube is:

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