Question

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

• A. $600\mathring{A}$
• B. $0.9\times {10}^{-12}m$
  
• C. $7\mathring{A}$
• D. $0.9\times {10}^{-19}nm$

Answer 1

The correct option is B $0.9\times {10}^{-12}m$

$eV=KE=\frac{1}{2}m{v}^2$

$KE=\frac{p^2}{2m}$

De-broglie wavelength,

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

$\lambda =\frac{h}{\sqrt{2mE}}=\frac{h}{\sqrt{2mVe}}$

Putting the values of constants, we get:

$\lambda =\frac{6.6\times {10}^{-34}}{\sqrt{2\times 1.6\times {10}^{-27}\times 1.6\times {10}^{-19}\times 1000}}=0.9\times {10}^{-12}m$