The wavelength of the de-Broglie wave associated with an electron accelerated through 150 V is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1 Å
An electron accelerated through a potential difference \( V \) gains kinetic energy \( eV \), so its momentum is \( p = \sqrt{2meV} \) and its de Broglie wavelength is \( \lambda = h/p \).
\( \displaystyle \lambda = \frac{h}{\sqrt{2meV}} \)
\( h = 6.63\times10^{-34}\,\mathrm{J\,s},\; m = 9.11\times10^{-31}\,\mathrm{kg},\; e = 1.6\times10^{-19}\,\mathrm{C} \)
\( \displaystyle \sqrt{2meV} = \sqrt{2 \times 9.11\times10^{-31} \times 1.6\times10^{-19} \times 150} = 6.61\times10^{-24}\,\mathrm{kg\,m\,s^{-1}} \)
\( \displaystyle \lambda = \frac{6.63\times10^{-34}}{6.61\times10^{-24}}\,\mathrm{m} = 1.0\times10^{-10}\,\mathrm{m} = 1\,\mathring{\mathrm{A}} \)
Shortcut: For an electron, putting in the constants gives the standard result \( \lambda = 12.27/\sqrt{V}\,\mathring{\mathrm{A}} \). Here \( \sqrt{150} = 12.25 \), so \( \lambda = 12.27/12.25 \approx 1\,\mathring{\mathrm{A}} \).