Question

The de Broglie wavelength of an electron accelerated by an electric field of $V$ volt is given by:

• A. $\lambda =\frac{1.23}{\sqrt{m}}$
• B. $\lambda =\frac{1.23m}{\sqrt{h}}m$
• C. $\lambda =\frac{1.23}{\sqrt{V}}nm$
• D. $\lambda =\frac{1.23}{V}$

Answer 1

Answer: (C)

$\lambda =\frac{1.23}{\sqrt{V}}nm$

The de Broglie wavelength of an electron accelerated by an electric field of V is $\frac{1.23}{\sqrt{V}}nm$
The kinetic energy of an electron accelerated by an electric field of V is $eV$. But it is also equal to $\frac{1}{2}m{u}^2$
$\frac{1}{2}m{u}^2=eV$
$u^2=\frac{2eV}{m}$
$u=\sqrt{\frac{2eV}{m}}$......(1)
The de Broglie wavelength $\lambda =\frac{h}{mu}$.....(2)
Substitute equation (1) into equation (2).
$\lambda =\frac{h}{m\times \sqrt{\frac{2eV}{m}}}$.
$\lambda =\frac{h}{\sqrt{2eVm}}$.
$\lambda =\frac{6.626\times {10}^{-34}Js}{\sqrt{2\times 1.6\times {10}^{-19}C\times V\times 9.1\times {10}^{-31}kg}}$.
$\lambda =\frac{6.626\times {10}^{-34}Js}{\sqrt{2\times 1.6\times {10}^{-19}C\times V\times 9.1\times {10}^{-31}kg}}$.
$\lambda =\frac{1.23\times {10}^{-9}}{\sqrt{V}}$ m.
$\lambda =\frac{1.23}{\sqrt{V}}$ nm.
Because 1 nm $={10}^{-9}$ m.