The de Broglie wavelength of an electron moving with a speed of \( 1.5\times10^{7}\,\mathrm{m\,s^{-1}} \) is equal to that of a photon. The ratio of the kinetic energy of the electron to the energy of the photon is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1/40
The de Broglie wavelength \( \lambda=h/p \) holds for the electron and the photon alike, so equal wavelengths mean equal momenta \( p \). The two energies are then written in terms of the same \( p \).
The kinetic energy of the electron, with \( p=mv \), is:
\( \begin{aligned} K &= \frac{p^{2}}{2m} \\ &= \frac{pv}{2} \end{aligned} \)
The energy of the photon is:
\( \begin{aligned} E &= \frac{hc}{\lambda} \\ &= pc \end{aligned} \)
With \( c=3\times10^{8}\,\mathrm{m\,s^{-1}} \), the ratio is:
\( \begin{aligned} \frac{K}{E} &= \frac{pv/2}{pc} \\ &= \frac{v}{2c} \\ &= \frac{1.5\times10^{7}\,\mathrm{m\,s^{-1}}}{2\times3\times10^{8}\,\mathrm{m\,s^{-1}}} \\ &= \frac{1}{40} \end{aligned} \)