An electron is accelerated to potential V. If mass of electron is \( m = 9.1 \times 10^{-31} \) kg its charge is \( e = 1.6 \times 10^{-19} \) C, then de-Broglie wavelength of electron is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \lambda = \displaystyle\frac{1.227}{\sqrt{v}}\,\mathrm{nm} \)
An electron accelerated through a potential \( V \) gains kinetic energy \( K = eV \). Its momentum is \( p = \sqrt{2mK} = \sqrt{2meV} \), and the de Broglie wavelength is \( \lambda = h/p \).
\( \displaystyle \lambda = \frac{h}{\sqrt{2meV}} \)
Taking \( h = 6.63\times10^{-34}\,\mathrm{J\,s} \):
\( \displaystyle \sqrt{2me} = \sqrt{2 \times 9.1\times10^{-31} \times 1.6\times10^{-19}} = 5.40\times10^{-25} \)
\( \displaystyle \lambda = \frac{6.63\times10^{-34}}{5.40\times10^{-25}\sqrt{V}}\,\mathrm{m} = \frac{1.227\times10^{-9}}{\sqrt{V}}\,\mathrm{m} = \frac{1.227}{\sqrt{V}}\,\mathrm{nm} \)
Here \( V \) is in volts; the options write it as \( v \).