An electron is accelerated through a potential difference of 36 V. The de-Broglie wavelength associated with it will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 2\times10^{-10}\,\mathrm{m} \)
An electron accelerated through \( V \) volts gains kinetic energy \( eV \), so its momentum is \( p = \sqrt{2meV} \) and its de Broglie wavelength is
\( \displaystyle \lambda = \frac{h}{\sqrt{2meV}} = \frac{1.227}{\sqrt{V}}\,\mathrm{nm} \)
\( \displaystyle \lambda = \frac{1.227}{\sqrt{36}}\,\mathrm{nm} = \frac{1.227}{6}\,\mathrm{nm} \approx 0.204\,\mathrm{nm} \)
\( \displaystyle \lambda \approx 2.04\times10^{-10}\,\mathrm{m} \approx 2\times10^{-10}\,\mathrm{m} \)