Question

A photon of wavelength 4 into 10 7 m strikes a metal surface the work function of the metal being 2.13 ev calculate the energy of the photon the kinetic energy of the emission

Answer 1

We are going to use Einstein's Explanation of the Photoelectric Effect.

In equation form, 

$W = \phi + K$

Where,

$W$ = Incident Energy
$\phi$ = Work Function
$K$ = Kinetic Energy


Here our data is:

$\lambda = 4\times 10^{-7} \, \, m \\ \\ \phi = 2.13 \, \, eV$

First, we need to calculate the energy of incident photons

$W = hf$

But we are given wavelength, so we use the relation:

$c=f\lambda \\ \\ \implies f = \frac{c}{\lambda}$

Hence we have:

$\displaystyle W = hf \\ \\ \\ \implies W = \frac{hc}{\lambda} \\ \\ \\ \implies W = \frac{6.626\times 10^{-34} \times 3 \times 10^8}{4\times 10^{-7}} \\ \\ \\ \implies W = 4.9695 \times 10^{-19} \, \, J \\ \\ \\ \left[\text{Now } \, 1 \, \, eV = 1.6 \times 10^{-19} \, \, J \right] \\ \\ \\ \implies W = \frac{4.9695\times 10^{-19}}{1.6 \times 10^{-19}} \, \, eV \\ \\ \\ \implies \boxed{\bold{W \approx 3.11 \, \, eV}}$


Thus, Energy of Incident Photons is approximately 3.11 eV



Now, we use Einstein's Equation to calculate Kinetic Energy of photoelectrons:

$W = \phi + K \\ \\ \\ \implies 3.11 = 2.13 + K \\ \\ \\ \implies K = 3.11-2.13 \\ \\ \\ \implies \boxed{\bold{K = 0.98 \, \, eV}} \\ \\ \\ \text{We can also convert it into joules} \\ \\ \\ \implies K = 0.98 \times 1.6 \times 10^{-19} \, \, J \\ \\ \\ \implies \boxed{\bold{K\approx 1.57 \times 10^{-19} \, \, J}}$


Thus, The Maximum Kinetic Energy of Photoelectrons is 0.98 eV or $\bold{1.57\times 10^{-19} \, \, J}$