Question

If the speed of photoelectrons is ${10}^{4}m/s$ , what should be the frequency of the incident radiation on a potassium metal ? Given, work function of potassium=$2.3eV$.

Answer 1

$E=W_0+E_k$ where, $W_0$ is the function and $E_k$ is the $KE$ of th liberated photelectron

$W_0=2.3eV=3.68\times {10}^{19}J$

$E_k=\frac{1}{2}m{v}^2$

$=\frac{1}{2}\times 9\times {10}^{-31}\times {\left({10}^4\right)}^{2}J=4.5\times {10}^{-23}J$

So ,$E=(3.68\times {10}^{-19}+4.5\times {10}^{-23})J$

$=3.68045\times {10}^{-19}J$

Frequency, $v=\frac{E}{h}=\frac{3.68045\times {10}^{19}}{(6.63\times {10}^{-34})}$

$=0.56\times {10}^{15}Hz$