An electron and a photon each have a wavelength of 1.00 nm. Find

Wavelength of an electron $\left(\lambda_{e}\right)$ and a photon $\left(\lambda_{p}\right), \lambda_{e}=\lambda_{p}=\lambda=1 \mathrm{~nm}$

= 1 × 10−9 m

Planck’s constant, h = 6.63 × 10−34 Js

(a) The momentum of an elementary particle is given by de Broglie relation:

$\lambda=\frac{h}{p}$

$p=\frac{h}{\lambda}$

It is clear that momentum depends only on the wavelength of the particle. Since the wavelengths of an electron and a photon are equal, both have an equal momentum.

$\therefore p=\frac{6.63 \times 10^{-34}}{1 \times 10^{-9}}=6.63 \times 10^{-25} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$

(b) The energy of a photon is given by the relation:

$E=\frac{h c}{\lambda}$

Where,

Speed of light, c = 3 × 108 m/s

$\therefore E=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{1 \times 10^{-9} \times 1.6 \times 10^{-19}}$

$=1243.1 \mathrm{eV}=1.243 \mathrm{keV}$

Therefore, the energy of the photon is 1.243 keV.

(c) The kinetic energy (K) of an electron having momentum p,is given by the relation:

$K=\frac{1}{2} \frac{p^{2}}{m}$

Where,

m = Mass of the electron = 9.1 × 10−31 kg

$p=6.63 \times 10^{-25} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$

$\therefore K=\frac{1}{2} \times \frac{\left(6.63 \times 10^{-25}\right)^{2}}{9.1 \times 10^{-31}}=2.415 \times 10^{-19} \mathrm{~J}$

$=\frac{2.415 \times 10^{-19}}{1.6 \times 10^{-19}}=1.51 \mathrm{eV}$

Hence, the kinetic energy of the electron is 1.51 eV.