The de-Broglie wavelength associated with an electron accelerated through a potential difference of 100 V is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.23 Å
The electron gains kinetic energy \( eV \) in the potential difference, and the de Broglie wavelength is \( \lambda = h/p \) with \( p = \sqrt{2meV} \).
\( \displaystyle \lambda = \frac{h}{\sqrt{2meV}} \)
\( \displaystyle \lambda = \frac{6.63\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times100}}\,\mathrm{m} \)
\( \displaystyle \lambda = \frac{6.63\times10^{-34}}{5.4\times10^{-24}}\,\mathrm{m} = 1.23\times10^{-10}\,\mathrm{m} = 1.23\,\mathrm{\mathring{A}} \)
Shortcut: For an electron accelerated through V volts, \( \lambda = 12.27/\sqrt{V}\,\mathrm{\mathring{A}} \). Putting \( V = 100 \) gives \( \lambda = 12.27/10 \approx 1.23\,\mathrm{\mathring{A}} \) directly.
Note: A similar question was also asked in CUET (UG) 2023 on 13-06-2023 (Slot 1).