Question

If the short wavelength limit of the continuous spectrum coming out of a Coolidge tube is $10\stackrel{o}{A}$, then the de Broglie wavelength of the electrons reaching the target metal in the Coolidge tube is approximately:

• A. $0.3\stackrel{o}{A}$
• B. $3\stackrel{o}{A}$
• C. $30\stackrel{o}{A}$
• D. $10\stackrel{o}{A}$

Answer 1

The correct option is A $0.3\stackrel{o}{A}$

$K.E=\frac{P^2}{2me}=\frac{hc}{{\lambda }_{min}}$

$\therefore P=\sqrt{\frac{2hcme}{{\lambda }_{min}}}$

Also

${\lambda }_{debroglie}=\frac{h}{p}=\sqrt{\frac{h{\lambda }_{min}}{2meC}}$

For

${\lambda }_{min}=10A°$

${\lambda }_{debroglie}=0.3A°$