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(C) According to Moseley's Law,the frequency $
u$ of the $K_\alpha$ line is given by $
u = c/\lambda = R(Z-1)^2(1/1^2 - 1/2^2)$.Thus,$1/\lambda \pro

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$(9/4) \lambda$

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(C) According to Moseley's Law,the frequency $
u$ of the $K_\alpha$ line is given by $
u = c/\lambda = R(Z-1)^2(1/1^2 - 1/2^2)$.Thus,$1/\lambda \propto (Z-1)^2$.Let $\lambda_1$ be the wavelength for $Z_1 = 43$ and $\lambda_2$ be the wavelength for $Z_2 = 29$.Then,$\lambda_2 / \lambda_1 = (Z_1 - 1)^2 / (Z_2 - 1)^2$.Substituting the values: $\lambda_2 / \lambda = (43 - 1)^2 / (29 - 1)^2$.$\lambda_2 / \lambda = (42)^2 / (28)^2 = (42/28)^2$.Simplifying the fraction $42/28 = 3/2$.Therefore,$\lambda_2 / \lambda = (3/2)^2 = 9/4$.Hence,$\lambda_2 = (9/4) \lambda$.

If the wavelength of the $K_\alpha$ line of an element with atomic number $43$ is $\lambda$,then the wavelength of the $K_\alpha$ line of an element with atomic number $29$ is:

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