The wavelength of light from the spectral emission line of sodium is 662 nm. The kinetic energy at which an electron would have same de-Broglie wavelength would be: \( (h = 6.62\times10^{-34}\,\mathrm{J\,s},\ m_e = 9\times10^{-31}\,\mathrm{kg}) \)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 5.5\times10^{-25}\,\mathrm{J} \)
The de Broglie wavelength of an electron is \( \lambda = h/p \), so the same wavelength means the same momentum. The kinetic energy then follows from \( K = p^{2}/2m \).
\( \lambda = 662\,\mathrm{nm} = 6.62\times10^{-7}\,\mathrm{m} \)
\( \displaystyle p = \frac{h}{\lambda} = \frac{6.62\times10^{-34}}{6.62\times10^{-7}} = 1\times10^{-27}\,\mathrm{kg\,m\,s^{-1}} \)
\( \displaystyle K = \frac{p^{2}}{2m_e} = \frac{(10^{-27})^{2}}{2 \times 9\times10^{-31}} = \frac{10^{-54}}{1.8\times10^{-30}} \)
\( \displaystyle K \approx 5.5\times10^{-25}\,\mathrm{J} \)