Question

Find the maximum magnitude of the linear momentum of a photoelectron emitted when a wavelength of 400 nm falls on a metal with work function 2.5 eV.

Answer 1

The maximum magnitude of the linear momentum of a photo electron emitted when a wavelength of 400 nm falls on a metal with work function 2.5 eV is $4.197 \times 10^{-25} \mathrm{kg}-\mathrm{m} / \mathrm{s}$

Explanation:

Given data in the question  

light Wavelength, λ = 400 nm = 400×10-9 m

metal Work function, ϕ = 2.5 eV

Step 1:

From the Photoelectric Equation of Einstein,

Kinetic energy $=\frac{h c}{\lambda}-\phi$

Here, c = light of velocity

          h = Planck's constant

$\begin{aligned}\therefore K \cdot E \cdot=\frac{\left(6.63 \times 10^{-34} \times 3 \times 10^{8}\right)}{\left(4 \times 10^{-7} \times 1.6 \times 10^{-19}\right)}-2.5 e V \\=0.605 \mathrm{eV}\end{aligned}$

Step 2:

Also,  $K . E =\frac{p^{2}}{2 m^{\prime}}$

Where p is momentum, and m is electron mass.

$\begin{array}{c}{\therefore p^{2}=2 m \times K . E} \\{p 2=2 \times 9.1 \times 10^{-31} \times 0.605 \times 1.6 \times 10^{-19}} \\{p=4.197 \times 10^{-25} \mathrm{kg}-m / s}\end{array}$