Question

The de-broglie wavelength of an electron accelerated by an electric potential of 'V' volts is given by
$(Given:m_e=9.1\times {10}^{-31}kg,e=1.6\times {10}^{-19}C,h=6.6\times {10}^{-34}J/s)$

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

Answer 1

The correct option is C $\lambda =\frac{1.23}{\sqrt{V}}nm$

From de-Broglie equation,
$\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2m{E}_k}}=\frac{h}{\sqrt{2meV}}$
$\lambda$ = wavelength
$E_k$ = kinetic energy
m = mass of the electron
h = planck's constant
Subtitute values in above equation,
$\lambda =\frac{6.6\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 1.6\times {10}^{-19}\times V}}\lambda =\frac{1.23}{\sqrt{V}}nm$