Physics - Modern Physics Question with Solution | TestHub

${\lambda }_{k\alpha }=0\text{.}021 \text{nm}=0\text{.}21\stackrel{\circ }{\text{A}}$

Since,${\lambda }_{k\alpha }$corresponds to the transition of an electron fromL-shell toK-shell, therefore

$E{}_L-E{}_K=\left(\text{in eV}\right)=\frac{12375}{\lambda \left(\mathrm{in} {\displaystyle \stackrel{\circ }{\mathrm{A}}}\right)}=\frac{12375}{0\text{.21}}\approx 589\mathrm{28 }\text{eV}$
or$\Delta E = 59 \text{keV}$