Question

The de-Broglie wavelength of a proton $(charge=1.6\times {10}^{-19}C,mass=1.6\times {10}^{-27}kg)$ accelerated through a potential difference of $1kV$ is :

• A. $600\stackrel{\circ }{A}$
• B. $0.9\times {10}^{-12}m$
• C. $7\stackrel{\circ }{A}$
• D. $0.9nm$

Answer 1

Answer: (B)

$0.9\times {10}^{-12}m$

According to de-Broglie hypothesis

$\lambda =\frac{h}{p}$
$=\frac{h}{\sqrt{2mE}}=\frac{h}{\sqrt{2mqV}}$

$\therefore \lambda =\frac{6.6\times {10}^{-34}}{\sqrt{2\times (1.6\times {10}^{-27})(1.6\times {10}^{-19})\times 1000}}$

$=\frac{6.6\times {10}^{-34}}{7.16\times {10}^{-22}}$

$=0.9\times {10}^{-12}m$