A metal target with atomic number Z=46 is bom | vedclass

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(A) The wavelength of the $K_\alpha$ line is given by Moseley's law: $\frac{1}{\lambda_{K_\alpha}} = R(Z-1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \ri

Vedclass · Accepted Answer

$2.53$

Vedclass · Answer

(A) The wavelength of the $K_\alpha$ line is given by Moseley's law: $\frac{1}{\lambda_{K_\alpha}} = R(Z-1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R(Z-1)^2 \left( \frac{3}{4} ight)$.The cut-off wavelength $\lambda_0$ is given by $\lambda_0 = \frac{hc}{eV}$,where $V$ is the accelerating potential of the electron beam.Given the ratio $r = \frac{\lambda_{K_\alpha}}{\lambda_0} = 2$ for $Z_1 = 46$.From the expression for $\lambda_{K_\alpha}$,we see that $\lambda_{K_\alpha} \propto \frac{1}{(Z-1)^2}$.Since the accelerating potential $V$ is the same,$\lambda_0$ remains constant. Therefore,$r \propto \frac{1}{(Z-1)^2}$.Thus,$\frac{r_2}{r_1} = \frac{(Z_1-1)^2}{(Z_2-1)^2}$.Substituting the values: $\frac{r_2}{2} = \frac{(46-1)^2}{(41-1)^2} = \frac{45^2}{40^2} = \left( \frac{9}{8} ight)^2 = \frac{81}{64} = 1.2656$.$r_2 = 2 	imes 1.2656 = 2.5312 \approx 2.53$.

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