The {K_{alpha}} X-rays arising from a cobalt | vedclass

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(B) According to Moseley's law, the wavelength of ${K_{\alpha}}$ $X$-rays is given by $\frac{1}{\lambda} = R(Z - 1)^2 \left( \frac{1}{1^2} - \frac{1}{

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$ < 179\, pm$

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(B) According to Moseley's law, the wavelength of ${K_{\alpha}}$ $X$-rays is given by $\frac{1}{\lambda} = R(Z - 1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} ight)$.This implies that $\lambda \propto \frac{1}{(Z - 1)^2}$.For cobalt $(Z_{Co} = 27)$ and nickel $(Z_{Ni} = 28)$, we have:$\frac{\lambda_{Ni}}{\lambda_{Co}} = \left( \frac{Z_{Co} - 1}{Z_{Ni} - 1} ight)^2 = \left( \frac{27 - 1}{28 - 1} ight)^2 = \left( \frac{26}{27} ight)^2$.Since $\frac{26}{27} Therefore, $\lambda_{Ni} Given $\lambda_{Co} = 179\, pm$, we find $\lambda_{Ni} = \left( \frac{26}{27} ight)^2 	imes 179 \approx 165.9\, pm$.Thus, the wavelength of ${K_{\alpha}}$ $X$-rays for nickel is $ < 179\, pm$.

The ${K_{\alpha}}$ $X$-rays arising from a cobalt $(Z = 27)$ target have a wavelength of $179\, pm$. The ${K_{\alpha}}$ $X$-rays arising from a nickel target $(Z = 28)$ is

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