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(C) According to Moseley's law for $K_\alpha$ $X$-rays,the frequency is given by $
u = c/\lambda = R c (Z-1)^2 (1/1^2 - 1/2^2)$.This simplifies to $\

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) According to Moseley's law for $K_\alpha$ $X$-rays,the frequency is given by $
u = c/\lambda = R c (Z-1)^2 (1/1^2 - 1/2^2)$.This simplifies to $\frac{1}{\lambda} = R(Z-1)^2 \left( \frac{3}{4} ight)$,where $R \approx 1.097 	imes 10^7 \ m^{-1}$ is the Rydberg constant.Rearranging for $Z$:$(Z-1)^2 = \frac{4}{3 R \lambda} = \frac{4}{3 	imes (1.097 	imes 10^7 \ m^{-1}) 	imes (0.76 	imes 10^{-10} \ m)}$.$(Z-1)^2 = \frac{4}{2.501 	imes 10^{-3}} \approx 1599.36$.Taking the square root,$Z-1 \approx 39.99 \approx 40$.Therefore,$Z \approx 41$. Given the options provided,$40$ is the closest integer value.

The wavelength of $K_\alpha$ $X$-rays produced in an $X$-ray tube is $0.76 \ \mathring{A}$. The atomic number of the anode material is:

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