An electron is accelerated through a potential difference of 144 V. The de-Broglie wavelength associated would be approximately:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.1 nm
An electron accelerated from rest through a potential difference V gains kinetic energy eV, so its momentum is \( p = \sqrt{2meV} \). Its de Broglie wavelength is
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}} = \frac{1.227}{\sqrt{V}}\,\mathrm{nm} \)
(the numerical form uses \( h = 6.63\times10^{-34}\,\mathrm{J\,s} \), \( m = 9.11\times10^{-31}\,\mathrm{kg} \), \( e = 1.6\times10^{-19}\,\mathrm{C} \), with V in volts).
\( \displaystyle \lambda = \frac{1.227}{\sqrt{144}}\,\mathrm{nm} = \frac{1.227}{12}\,\mathrm{nm} \approx 0.1\,\mathrm{nm} \)
Note: A similar question was also asked in CUET (UG) 2023 on 20-06-2023 (Slot 3).