Question

If the frequency of $K_{\alpha },K_{\beta }\text{and}{L}_{\alpha }X-rays$ for a material are $v_{K_{\alpha }},v_{K_{\beta }},v_{L_{\alpha }}$ respectively, then

• A. $v_{K_{\alpha }}=v_{K_{\beta }}+v_{L_{\alpha }}$
• B. $v_{L_{\alpha }}=v_{K_{\alpha }}+v_{K_{\beta }}$
• C. $v_{K_{\beta }}=v_{K_{\alpha }}+v_{L_{\alpha }}$
• D. None of these

Answer 1

The correct option is C $v_{K_{\beta }}=v_{K_{\alpha }}+v_{L_{\alpha }}$

The $K_{\alpha }$ ​ is emitted when the atom transitions from $K$ energy state to the $L$ energy state.
$E_L-E_K=h{\nu }_{K_{\alpha }}....\left(i\right)$
The $K_{\beta }$ is emitted when the atom transitions from $K$ energy state to the $M$ energy state.
$E_M-E_K=h{\nu }_{K_{\beta }}....\left(ii\right)$
The $L_{\alpha }$ ​ is emitted when the atom transitions from $L$ energy state to the $M$ energy state.
$E_M-E_L=h{\nu }_{L_{\alpha }}....\left(ii\right)$
So balancing energies, Energy released in $K_{\beta }$ ​ emission is equal to sum of those released in $K_{\alpha }$ ​ and $L_{\alpha }$ ​ emissions.
$h{v}_{K_{\alpha }}+h{v}_{L_{\alpha }}=h{v}_{K_{\beta }}[\because E=hv]$
$v_{K_{\beta }}=v_{K_{\alpha }}+v_{L_{\alpha }}$