Question

If the de-Broglie wavelength of a proton is $ {{10}^{-13}}m, $ the electric potential through which it must have been accelerated is:

• A. $ 4.07\times {{10}^{4}}\,V $
• B. $ 8.2\times {{10}^{4}}\,V $
• C. $ 8.2\times {{10}^{3}}\,V $
• D. $ 4.07\times {{10}^{5}}\,V $

Answer 1

Answer: (B)

$\lambda=\frac{h}{\sqrt{2 m q V}} .$
$\left(10^{-13}\right)^{2}=\frac{\left(6.6 \times 10^{-34}\right)^{2}}{2 \times 1.67 \times 10^{-27} \times 1.6 \times 10^{-19} V}$
$V=8.2 \times 10^{4} Volt .$