Question

A light of frequency 2.5ร10^15 hz is incident on a metal surface having work function 4 ev the velocity of photoelectron is

Answer 1

Answer:  $\textbf{1.5 \times 10^6 m/s}$


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


$\boxed{W = \phi + K} \\ \\ \\ OR \: \: \boxed{hf =hf_{\circ}+K}$

A light of some energy greater than the threshold Energy falls on Metal Surface. The energy equal to Threshold Energy is used up in emitting of Photoelectrons. The extra energy left is converted into the Kinetic Energy of Photoelectrons.


Let us calculate Incident Energy:

$\text{Incident Energy} = W = hf \\ \\ \\ \implies W = \text{Planck's Constant} \times \text{Incident Radiation Frequency} \\ \\ \\ \implies W = (6.626\times 10^{-34}) \times (2.5 \times 10^{15}) \, \, J \\ \\ \\ \implies W = 1.6565 \times 10^{-18} \, \, J \\ \\ \\ Now, \, \, 1 \, \, eV = 1.6 \times 10^{-19} \, \, J \\ \\ \\ \implies W = \frac{1.6565 \times 10^{-18}}{1.6 \times 10^{-19}} \, \, eV \\ \\ \\ \implies W = 10.35 \, \, eV$

Now, we have the Threshold Energy as:

$\phi = 4 \, \, eV$
From Einstein's equation:

$W = \phi + K \\ \\ \implies 10.35 = 4 + K \\ \\ \implies K = 6.35 \, \, eV$

Now, we have the Kinetic Energy of the Electron in terms of Electron Volts. We can convert it into joules and then find the velocity of electron:

$K=6.35\,\, eV\\ \\ \\ \left[Now \,\, 1\,\, eV=1.6\times 10^{-19}\,\, J\right]\\ \\ \\ \implies K=6.35\times 1.6\times 10^{-19}\,\, J \\ \\ \\ \implies\frac{1}{2}mv^2=1.016 \times 10^{-18} \, \, J \\ \\ \\ \implies v^2 = \frac{2\times 1.016 \times 10^{-18}}{m} \\ \\ \\ \implies v^2 = \frac{2.032 \times 10^{-18}}{9.1 \times 10^{-31}} \\ \\ \\ \implies v^2 =2.33 \times 10^{12} \\ \\ \\ \implies v \approx 1.49 \times 10^6 \, \, m/s \\ \\ \\ \implies \boxed{v\approx 1.5 \times 10^6 \,\, m/s}$

Answer 2

Answer:

The velocity of photo electron is $1.5\times10^{6}\ m/s$.

Explanation:

Given that,

Frequency $f= 2.5\times10^{15}\ Hz$

Work function $W=4 ev$

We know that,

Einstein explanation of the photoelectric effect is

$W= \phi+k$

Now,

Incident energy is,

$W=hf$

$W=6.625\times10^{-34}\times2.5\times10^{15}\ J$

$W=1.656\times10^{-18}\ J$

We know that,

$1\ ev=1.6\times10^{-19}\ J$

$W=\dfrac{1.656\times10^{-18}}{1.6\times10^{-19}}$

$W= 10.35\ ev$

We have threshold frequency

$\phi=4\ ev$

From Einstein equation,

$W=\phi+k$

$10.35=4+k$

$k=6.35\ ev$

$k= 6.35\times1.6\times10^{-19}\ J$

$k=1.016\times10^{-18}$

Here, K = kinetic energy

Therefore,

$\dfrac{1}{2}mv^{2}=1.016\times10^{-18}$

$\dfrac{1}{2}\times9.1\times10^{-31}\times v^{2}=1.016\times10^{-18}$

$v^{2}=\dfrac{2\times1.016\times10^{-18}}{9.1\times10^{-31}}$

$v^{2}=2.33\times10^{12}$

$v= 1.5\times10^{6}\ m/s$

Hence, The velocity of photo electron is $1.5\times10^{6}\ m/s$.