Question

In a coolidge tube, the potential difference used to accelerate the electrons is increased from $12.4\text{kV}$ to $24.8\text{kV}$. As a result, the difference between the wavelength of $K_{\alpha }$ -line and minimum wavelength increases two-fold. The wavelength (in $\AA$) of the $K_{\alpha }$ line is $(\frac{hc}{e}=12.4\text{kVA})$

Answer 1

Cut-off wavelength of continuous $x-$Ray spectrum is :-
${\lambda }_{\text{min}}=\frac{hc}{e\Delta V}$
Hence, $\frac{hc}{e}=12.4\text{KV}\AA$
${\lambda }_{\text{min}}=\frac{12.4\text{KV}\AA }{\Delta V}$
Where $\Delta V=$ potential - difference.
For $\Delta V=12.4\text{KV}$:-
${\lambda }_{{\text{min}}_1}=\frac{12.4\text{KV}\AA }{12.4\text{KV}}=1\AA$
Similarly, for $\Delta V=24.8\text{KV}$:-
${\lambda }_{{\text{min}}_2}=\frac{12.4\text{KV}\AA }{24.8\text{KV}}=0.5\AA$
$\to$ Let, $\lambda$ be the wavelength of $K_{\alpha }$ line which is independent of accelerating potential.
Given:-
$[\lambda -{\lambda }_{{\text{min}}_2}]=2[\lambda -{\lambda }_{{\text{min}}_1}]$
$[\lambda -0.5]=2[\lambda -1]$
$\lambda -0.5=2\lambda -2$
$\lambda =1.5\AA$