Question

If the wavelength of a photon and the wavelength of an an electron are both 250 nm, calculate the ratio of the photon energy to the electron energy.   Which has the greater energy?

Answer 1

The energy of a photon having wavelength \(\lambda\) is

\(E=\frac{h c}{\lambda}\)

\(\mathrm{h}=6.63^{*} 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} / \mathrm{s}\)

\(\mathrm{c}=3^{*} 10^{8} \mathrm{~m} / \mathrm{s}\)

\(\lambda=250 \mathrm{~nm}=250^{*} 10^{-9} \mathrm{~m}\)

\(E=\frac{6.63 * 10^{-34} * 3 * 10^{8}}{250 * 10^{-9}}\)

\(E=7.956 * 10^{-19} J\) (Ans)

For an electron of wavelength \(\lambda\), the wavelength is defined as

\(\lambda=\frac{h}{m v}\)

Therefore, its velocity is

\(v=\frac{h}{m \lambda}\)

\(\mathrm{h}=6.63^{*} 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} / \mathrm{s}\)

\(\mathrm{m}=9.1^{*} 10^{-31} \mathrm{~kg}\)

\(\lambda=250 \mathrm{~nm}=250^{*} 10^{-9} \mathrm{~m}\)

\(v=\frac{6.63 * 10^{-34}}{9.1 * 10^{-31} * 250 * 10^{-9}}\)

\(v=2.914 * 10^{3} \mathrm{~m} / \mathrm{s}\)

The energy of moving electron is

\(E=\frac{1}{2} m v^{2}\)

\(E=\frac{1}{2} * 9.1 * 10^{-31} *\left(2.914 * 10^{3}\right)^{2}\)

\(E=3.86 * 10^{-24} J\) (Ans)