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 $1 \,kV$ is

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

Answer 1

Answer: (B)

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 $