Question

Find the maximum magnitude of the linear momentum of a photo electron emitted when light of wavelength 400 nm falls on a metal having work function 2.5 eV.

Answer 1

Given,

Wave length of the light = 400 nm = $4\times {10}^{-7}m$

We know,

Kinetic energy of Photo electron :

$\Rightarrow \frac{1}{2}m{v}^2=\frac{hc}{\lambda }-h{v}_1$

$=\frac{4.14\times {10}^{-15}\times 3\times {10}^8}{4\times {10}^{-7}}-2.5eV$

$=0.605eV$

Hence, the kinetic energy of the photo electron is 0.605 eV

We know,

KE = $\frac{P^2}{2m}$

Or we can also write

$\Rightarrow P^2=2m\times \text{KE}$

Putting all the values

$P^2=2\times 9.1\times {10}^{-31}\times 0.605\times 1.6\times {10}^{-19}$

$P=4.197\times {10}^{-25}kg-m/s$

Hence, the maximum magnitude of linear momentum is $P=4.197\times {10}^{-25}kg-m/s$