Question

For a certain metal, the $K$ absorption edge is at $0.172\mathring{A}$. The wavelength of $K_{\alpha },K_{\beta }$ and $K_{\gamma }$ lines of $K$ series are $0.210\mathring{A},0.192\mathring{A}$ and $0.108\mathring{A}$ respectively. The energies of $K,L$ and $M$ orbits are $E_K,E_L$ and $E_M$, respectively. Then

• A. $E_K=-13.04keV$
• B. $E_L=-7.52keV$
• C. $E_M=-3.21keV$
• D. $E_K=13.04keV$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $E_K=-13.04keV$
B $E_M=-3.21keV$
C $E_L=-7.52keV$

Energy corresponding to K absorption edge=$\frac{hc}{{\lambda }_K}$

Energy of $K_n$ line of K series=$\frac{hc}{{\lambda }_{K_n}}$

Thus energy of $K$ orbit, $E_K=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_{K_{\alpha }}})=-13.04keV$

Energy of $L$ orbit, $E_L=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_{K_{\beta }}})=-7.52keV$

Energy of $M$ orbit, $E_M=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_{K_{\gamma }}})=-3.21keV$