The de Broglie wavelength of an electron having kinetic energy 56 eV is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.164 nm
The de Broglie wavelength is \( \lambda = h/p \), and a particle of kinetic energy K has momentum \( p = \sqrt{2mK} \).
\( K = 56\,\mathrm{eV} = 56\times1.6\times10^{-19}\,\mathrm{J} = 8.96\times10^{-18}\,\mathrm{J} \)
\( p = \sqrt{2\times9.1\times10^{-31}\times8.96\times10^{-18}}\,\mathrm{kg\,m\,s^{-1}} = 4.04\times10^{-24}\,\mathrm{kg\,m\,s^{-1}} \)
\( \displaystyle \lambda = \frac{h}{p} = \frac{6.63\times10^{-34}}{4.04\times10^{-24}}\,\mathrm{m} = 1.64\times10^{-10}\,\mathrm{m} = 0.164\,\mathrm{nm} \)
Shortcut: For an electron accelerated through V volts, \( \lambda = 12.27/\sqrt{V}\ \mathrm{\mathring{A}} \). An electron of 56 eV corresponds to \( V = 56 \), so \( \lambda = 12.27/7.48 = 1.64\ \mathrm{\mathring{A}} = 0.164\,\mathrm{nm} \).