What is the de-Broglie wavelength associated with an electron, accelerated through a potential difference of 10 kV?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 0.0123\,\mathrm{nm} \)
An electron accelerated through a potential difference \( V \) gains kinetic energy \( eV \), so its momentum is \( p = \sqrt{2meV} \). The de Broglie wavelength is
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}} \)
\( h = 6.63\times10^{-34}\,\mathrm{J\,s} \)
\( m = 9.1\times10^{-31}\,\mathrm{kg} \)
\( e = 1.6\times10^{-19}\,\mathrm{C} \)
\( V = 10\,\mathrm{kV} = 10^{4}\,\mathrm{V} \)
\( \displaystyle \sqrt{2meV} = \sqrt{2 \times 9.1\times10^{-31} \times 1.6\times10^{-19} \times 10^{4}} = 5.4\times10^{-23}\,\mathrm{kg\,m\,s^{-1}} \)
\( \displaystyle \lambda = \frac{6.63\times10^{-34}}{5.4\times10^{-23}} = 1.23\times10^{-11}\,\mathrm{m} = 0.0123\,\mathrm{nm} \)
Shortcut: For an electron, putting the constants in once gives the standard result
\( \displaystyle \lambda = \frac{1.227}{\sqrt{V}}\,\mathrm{nm} = \frac{1.227}{\sqrt{10^{4}}}\,\mathrm{nm} = \frac{1.227}{100}\,\mathrm{nm} = 0.0123\,\mathrm{nm} \)