Question

The $ K _{\alpha }$ $X$ -ray emission line of tungsten occurs at $\lambda = 0 \text{.} 0 2 1 \, \text{nm} \text{.}$ The energy difference between $K$ and $L$ levels in this atom is about

• A. $0 . 5 1 \text{ MeV}$
• B. $1. 2 \text{ MeV}$
• C. $5 9 \text{ keV}$
• D. $1 3 \text{.} 6 \text{ eV}$

Answer 1

Answer: (C)

$\lambda _{k \alpha }=0\text{.}021 \, \text{nm}=0\text{.}21\overset{^\circ }{\text{A}}$
Since, $\lambda _{ k \alpha }$ corresponds to the transition of an electron from L-shell to K-shell, therefore
$E _{L ⁡}-E⁡_{K ⁡}=\left(\text{in eV}\right)=\frac{1 2 3 7 5}{\lambda \left(in \, \overset{^\circ }{A}\right)}=\frac{1 2 3 7 5}{0 \text{.21}}\approx 58928 \, \text{eV}$
or $\Delta E \, = \, 59 \, \text{keV}$