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(B) In an $X$-ray spectrum,the $K$-series lines are produced when electrons transition from higher energy levels $(n=2, 3, 4, ...)$ to the $K$-shell $

Vedclass · Accepted Answer

$\lambda_\alpha > \lambda_\beta > \lambda_\gamma$

Vedclass · Answer

(B) In an $X$-ray spectrum,the $K$-series lines are produced when electrons transition from higher energy levels $(n=2, 3, 4, ...)$ to the $K$-shell $(n=1)$.The energy of the emitted photon is given by $E = \frac{hc}{\lambda}$,which implies $\lambda = \frac{hc}{E}$.The energy difference for the transitions are:For $K_\alpha$: $E_\alpha = E_2 - E_1$For $K_\beta$: $E_\beta = E_3 - E_1$For $K_\gamma$: $E_\gamma = E_4 - E_1$Since $E_4 > E_3 > E_2$,it follows that $E_\gamma > E_\beta > E_\alpha$.Because wavelength is inversely proportional to energy,we have $\lambda_\alpha > \lambda_\beta > \lambda_\gamma$.

In an $X$-ray spectrum,if the wavelengths of $K_\alpha$,$K_\beta$,and $K_\gamma$ lines are $\lambda_\alpha$,$\lambda_\beta$,and $\lambda_\gamma$ respectively,then:

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