Question

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

• A. $600\mathring{\text{A}}$
• B. $0.9\times {10}^{-12}\text{m}$
• C. $7\mathring{\text{A}}$
• D. $9\mathring{\text{A}}$

Answer 1

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

The de-Broglie wavelength is given by,

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

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

$=\frac{6.6\times {10}^{-34}\times {10}^{22}}{1.6\sqrt{20}}=0.9\times {10}^{-12}\text{m}.$

Hence, $\left(B\right)$ is the correct answer.