Question

In Coolidge tube of original voltage 20 kV, if voltage is increased by 10 kV, then change in cut-off wavelength is [take $hc$ = 1240 eV-nm]

• A. 0 pm
• B. 20.67 pm
• C. 60.3 pm
• D. 80 pm

Answer 1

The correct option is B 20.67 pm

$\lambda =\frac{hc}{eV}\Rightarrow {\lambda }_1=\frac{hc\times {10}^{-3}}{e\times 20}$

and ${\lambda }_2=\frac{hc\times {10}^{-3}}{e\times 30}$

${\lambda }_1-{\lambda }_2=\frac{hc}{e}\times {10}^{-3}(\frac{1}{20}-\frac{1}{30})=\Rightarrow \frac{1240\times {10}^{-12}}{60}m=20.67pm$