Moseley's law for characteristic X-rays is gi | vedclass

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(B) Moseley's law relates the frequency $f$ of characteristic $X$-rays to the atomic number $Z$ of the target material as $\sqrt{f} = a(Z - b)$.For $K

Vedclass · Accepted Answer

Both $a$ and $b$ are independent of the target material.

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(B) Moseley's law relates the frequency $f$ of characteristic $X$-rays to the atomic number $Z$ of the target material as $\sqrt{f} = a(Z - b)$.For $K_{\alpha}$ lines,the frequency is given by $f = \frac{m_e q_e^4}{8 h^3 \varepsilon_0^2} \left( \frac{3}{4} \right) (Z - 1)^2$.Taking the square root,we get $\sqrt{f} = \sqrt{\frac{3 m_e q_e^4}{32 h^3 \varepsilon_0^2}} (Z - 1)$.Comparing this with $\sqrt{f} = a(Z - b)$,we identify $a = \sqrt{\frac{3 m_e q_e^4}{32 h^3 \varepsilon_0^2}}$ and $b = 1$.Here,$a$ is a constant involving fundamental physical constants (mass of electron $m_e$,charge $q_e$,Planck's constant $h$,and permittivity $\varepsilon_0$),and $b$ is the screening constant (which is $1$ for $K_{\alpha}$ lines).Both $a$ and $b$ are independent of the target material $Z$.

Moseley's law for characteristic $X$-rays is given by $\sqrt{f} = a(Z - b)$. Choose the correct statement.

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