Question

The de-Broglie wavelength of a proton accelerated by $400V$ is

• A. $0.005\mathring{A}$
• B. $1.0528\mathring{A}$
• C. $0.0568\mathring{A}$
• D. $0.0143\mathring{A}$

Answer 1

The correct option is D $0.0143\mathring{A}$

$0.043\dot{A}$

formulae,

$qv=\frac{1}{2}m{v}^2$

$\Rightarrow v=\sqrt{\frac{2qV}{m}}$

$\Rightarrow v=\sqrt{\frac{2\times 1.6\times {10}^{-19}\times 400}{1.67\times {10}^{-27}}}=2.77\times {10}^{5}m/s$

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

$=\frac{6.63\times {10}^{-34}}{(1.67\times {10}^{-27})(2.77\times {10}^5)}$

$\therefore \lambda =1.43\times {10}^{-12}m$

$\Rightarrow \lambda =0.0143\dot{A}$