Question

A) An electron and a photon each have a wavelength of 1$.00\text{nm}$. Find their momenta.

B) An electron and a photon each have a wavelength of $1.00\text{nm}$. Find the energy of the photon.

C) An electron and a photon each have a wavelength of $1.00\text{nm}$. Find the kinetic energy of electron. (Momentum of the electron $=6.6\times {10}^{-25}\text{kg m/s})$

Answer 1

Formula used:$p=\frac{h}{\lambda }$
Given,
wavelength of electron and photon, $\lambda =1\text{nm}=1\times {10}^{-9}m$
Momentum of elementary particle is
given by de Broglie relation, $p=\frac{h}{\lambda }$
As the momentum depends only on the wavelength of the particle. Since the wavelength of the electron and photon are same so both have equal momentum,
$p=\frac{6.6\times {10}^{-34}}{1\times {10}^{-9}}$
$p=6.6\times {10}^{-25}\text{kgm/s}$

B) Formula used: $E=\frac{hc}{\lambda }$
Given, wavelength of electron and photon, $\lambda =1\text{nm}=1\times {10}^{-9}m$
Energy of the photon, $E=\frac{hc}{\lambda }$
Here, h=plank constant $=6.6\times {10}^{-34}Js$
c = speed of lligth $=3\times {10}^{8}m/s$
$E=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{1\times {10}^{-9}}$
$E=1.99\times {10}^{-16}J$
$E=\frac{1.99\times {10}^{-16}}{1.6\times {10}^{-19}}eV$
$E=1.24\times {10}^{3}eV=1.24\text{keV}$
Final answer : $1.24\text{keV}$

C) Formula used:
$K=\frac{p^2}{2m}$
Given, momentum of the electron $=6.6\times {10}^{-25}\text{kg m/s}$
Kinetic energy of the electron, $K=\frac{p^2}{2m}$
Here, 𝑚=mass of electron $=9.1\times {10}^{-31}\text{kg}$
$K=\frac{(6.6\times {10}^{-25}{)}^2}{2\times 9.1\times {10}^{-31}}$
$K=2.41\times {10}^{-19}J$
$K=\frac{2.41\times {10}^{-19}}{1.6\times {10}^{-19}}$
$K=1.51eV$
Final answer : $1.51eV$