The de Broglie wavelength associated with an electron accelerated through a potential difference of 100 V will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.123 nm
An electron accelerated through a potential difference V gains kinetic energy \( eV \), so its momentum is \( p = \sqrt{2meV} \), and the de Broglie wavelength is \( \lambda = h/p \).
\( \displaystyle \lambda = \frac{h}{\sqrt{2meV}} = \frac{6.63\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times100}} \)
\( \lambda \approx 1.23\times10^{-10}\,\mathrm{m} = 0.123\,\mathrm{nm} \)
Shortcut: For an electron, \( \lambda\,(\mathrm{\mathring{A}}) = 12.27/\sqrt{V} \) with V in volts. For \( V = 100 \) this is \( 12.27/10 = 1.227\,\mathrm{\mathring{A}} = 0.123\,\mathrm{nm} \).
Note: A similar question was also asked in CUET (UG) 2023 on 02-06-2023 (Slot 2).