What is the de Broglie wavelength associated with an electron accelerated through a potential difference of 144 V?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.102 nm
The de Broglie wavelength of a particle is \( \lambda = h/p \). An electron accelerated from rest through a potential difference \( V \) gains kinetic energy \( eV \), so its momentum is \( p = \sqrt{2meV} \).
\( \displaystyle \lambda = \frac{h}{\sqrt{2meV}} \)
Putting in \( h = 6.63\times10^{-34}\,\mathrm{J\,s} \), \( m = 9.1\times10^{-31}\,\mathrm{kg} \) and \( e = 1.6\times10^{-19}\,\mathrm{C} \) gives the standard result for electrons:
\( \displaystyle \lambda = \frac{1.227}{\sqrt{V}}\,\mathrm{nm} \)
\( \displaystyle \lambda = \frac{1.227}{\sqrt{144}}\,\mathrm{nm} = \frac{1.227}{12}\,\mathrm{nm} = 0.102\,\mathrm{nm} \)
Note: A similar question was also asked in CUET (UG) 2023 on 31-05-2023 (Slot 2).
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), but it is still in the CUET 2026 syllabus, so do not skip it.
NCERT (Pre-rationalised edition, 2022-23):
Wave Nature of Matter – “… Consider an electron (mass m, charge e) accelerated from rest through a potential V. … The de Broglie wavelength λ of the electron is then
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2meV}} \qquad (11.10) \)
Substituting the numerical values of h, m, e, we get
\( \displaystyle \lambda = \frac{1.227}{\sqrt{V}}\ \mathrm{nm} \qquad (11.11) \)
where V is the magnitude of accelerating potential in volts.”