Question

An electron wave has wavelength $1\mathring{A}$. Calculate the potential with which the electron is accelerated.

• A. $12.26$ volt.
• B. $150$ volt.
• C. $0.0826$ volt.
• D. $0.066$ volt.

Answer 1

The correct option is C $0.0826$ volt.

$\frac{1}{2}m{v}^2=eV,$ where $V=$ Potential with which electron is accelarated.

$\therefore mv=\sqrt{2meV}$

$\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2meV}}$ [de-broglie wavelength.]

$\therefore V=\frac{h^2}{2me{\lambda }^2}=\frac{{(6.6\times {10}^{-34})}^2}{2\times 9.1\times {10}^{-31}\times 1.6\times {10}^{-19}\times {\left({10}^{-10}\right)}^2}\approx 150volt$