Question

The ratio of the energy of the photon emitted by the $k_{\alpha }$ line to that of the $k_{\beta }$ line is :

• A. greater than $1$
• B. less than $1$
• C. $1$
• D. indeterminate

Answer 1

The correct option is B less than $1$

$E_{\alpha }=-13.6Z^2(\frac{1}{1^2}-\frac{1}{2^2})=-13.6Z^2\frac{3}{4}$

$E_{\beta }=-13.6Z^2(\frac{1}{1^2}-\frac{1}{3^2})=-13.6Z^2\frac{8}{9}$

$\frac{E_{\alpha }}{E_{\beta }}=\frac{3}{4}\times \frac{9}{8}=\frac{27}{32}$, which is less than $1$.