Question

If the maximum speed of the photoelectron emitted from a potassium surface $(\varphi =2.3\text{eV})$ is ${10}^6{\text{ms}}^{-1}$. What should be the frequency of the incident radiation? $(h=6.63\times {10}^{-34}\text{Joule-sec and}{m}_{\text{electron}}=9\times {10}^{-31}\text{kg})$

• A. $0.56\times {10}^{12}\text{Hz}$
• B. $0.56\times {10}^{15}\text{Hz}$
• C. $0.56\times {10}^{19}\text{Hz}$
• D. $1.23\times {10}^{15}\text{Hz}$

Answer 1

The correct option is D $1.23\times {10}^{15}\text{Hz}$

According to Einstein's photoelectric equation,

$E=K_{max}+\varphi =\frac{1}{2}m{v}_{max}^2+\varphi$

$=(\frac{1}{2}\times 9\times {10}^{-31}\times {\left({10}^6\right)}^2)+(2.3\times 1.6\times {10}^{-19})$

$=8.18\times {10}^{-19}\text{J}$

Energy of the photon, $E=h\nu$

$\Rightarrow \nu =\frac{E}{h}=\frac{8.18\times {10}^{-19}}{6.63\times {10}^{-34}}\text{Hz}$

$\Rightarrow \nu =1.23\times {10}^{15}\text{Hz}$

Hence, option $\left(D\right)$ is correct.