Question

A light source of frequency 6*10^15 ha is subjected to a metal sheet having work function equal to 22.6eV. Then the velocity of photo electrons emitted having maximum kinetic energy?

Answer 1

Let me know it you find an error in this.

Answer 2

Answer:

The velocity of photo electrons emitted is $\bold{2.8 \times 10^{-10} m/s}$

Explanation:

Given:  

A light source has a frequency of $6 \times 10^{15}$, then this light source is subjected to the metal sheet which has a work function which is equal to 22.6eV.

We have to find the velocity of photo electrons which are having maximum amount of kinetic energy.

That is,

$\bold{\vartheta=6 \times 10^{15}, \emptyset=22.6 \mathrm{eV}}$

We know that,

E=∅+K

Where $\mathrm{E}=\mathrm{h} \vartheta, \mathrm{k}=\frac{1}{2} m v^{2}$

Therefore,

$\bold{\mathrm{h} \vartheta=\varnothing+\frac{1}{2} m v^{2}}$ ………………eqn1

Where h= Planck’s constant= 6.63 \times 10^{-34}

$\bold{m=9.1 \times 10^{-31}}$

Now by substituting the known values in eqn1

$\begin{array}{l}{\left(6.63 \times 10^{-34}\right)\left(6 \times 10^{15}\right)=\left(22.6 \times 1.6 \times 10^{-19}\right)+\frac{1}{2} \times\left(9.1 \times 10^{-31} \times v^{2}\right)} \\ {=39.78 \times 10^{-19}=36.16 \times 10^{-91}+4.55 v^{2}} \\ {=3.62 \times 10^{-19}=4.55 v^{2}}\end{array}$

$\begin{array}{l}{V^{2}=\frac{3.62 \times 10^{-19}}{4.55}} \\ {V^{2}=0.795 \times 10^{-19}}\end{array}$

$\begin{array}{l}{V^{2}=7.95 \times 10^{-20}} \\ {V=\sqrt{7.95 \times 10^{-20}}}\end{array}$

Approx = $2.8 \times 10^{-10} m/s$

Explanation: