Question

A proton is accelerated to $225V$. Its de-Broglie wavelength is:

• A. $0.0019nm$
• B. $0.02nm$
• C. $0.003nm$
• D. $0.4nm$

Answer 1

The correct option is A $0.0019nm$

Energy of proton = $\frac{p^2}{2m}=qV$

$\frac{p^2}{2m}=225eV$

$p=\sqrt{2m\times 225\times 1.6\times {10}^{-19}}$

$\therefore \lambda =\frac{h}{p}$

$=\frac{6.626\times {10}^{-34}}{\sqrt{2m\times 225\times 1.6\times {10}^{-19}}}$

$=0.19\times {10}^{-11}m$

$=1.9\times {10}^{-12}m$

$=1.9pm$ or $0.001nm$