Question

If the operating potential in an X-ray tube is increased by 1 %, by what percentage does the cutoff wavelength decrease ?

Answer 1

We know $\lambda =\frac{hc}{V}$

Now, ${\lambda }^{\prime }=\frac{hc}{1.01V}=\frac{\lambda }{1.01}$

${\lambda }^{\prime }-\lambda =\frac{0.01}{1.01}\lambda$

% change of wave length

= $\frac{0.01\lambda }{1.01\times \lambda }\times 100=\frac{1}{01}$

= 0.9900

= 1 % approximately