In an X-ray tube,when the accelerating voltag | vedclass

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(A) Let the initial wavelength of the $K_\alpha$ line be $\lambda_{K_\alpha}$ and the initial minimum wavelength be $\lambda_{\min} = \frac{12400}{100

Vedclass · Accepted Answer

$29$

Vedclass · Answer

(A) Let the initial wavelength of the $K_\alpha$ line be $\lambda_{K_\alpha}$ and the initial minimum wavelength be $\lambda_{\min} = \frac{12400}{10000} \ \mathring{A} = 1.24 \ \mathring{A}$.After increasing the voltage to $20 \ kV$,the new minimum wavelength is $\lambda_{\min}' = \frac{12400}{20000} \ \mathring{A} = 0.62 \ \mathring{A}$.The problem states that the difference $(\lambda_{K_\alpha} - \lambda_{\min}')$ is $3$ times the difference $(\lambda_{K_\alpha} - \lambda_{\min})$.So,$\lambda_{K_\alpha} - \lambda_{\min}' = 3(\lambda_{K_\alpha} - \lambda_{\min})$.Rearranging gives $2\lambda_{K_\alpha} = 3\lambda_{\min} - \lambda_{\min}'$.Using Moseley's law,$\lambda_{K_\alpha} = \frac{4}{3R(Z-1)^2}$,where $\frac{1}{R} \approx 912 \ \mathring{A}$.Substituting the values: $2 \left[ \frac{4 	imes 912}{3(Z-1)^2} ight] = 3(1.24) - 0.62 = 3.72 - 0.62 = 3.10 \ \mathring{A}$.$\frac{8 	imes 304}{(Z-1)^2} = 3.10 \implies (Z-1)^2 = \frac{2432}{3.10} \approx 784.5$.Taking the square root,$Z-1 \approx 28$,so $Z = 29$.

In an $X$-ray tube,when the accelerating voltage is increased from $10 \ kV$ to $20 \ kV$,the difference between the wavelength of the $K_\alpha$ line and the short-wavelength limit of the continuous spectrum increases by a factor of $3$. The atomic number of the target element is:

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