Question

Calculate the maximum kinetic energy (in eV) of the emitted photoelectrons.

• A. 1.51
• B. 2.36
• C. 3.85
• D. 4.27

Answer 1

The correct option is A 1.51

Given , threshold frequency ${\lambda }_0=3800A^o=3800\times {10}^{-10}m$ and wavelength of incident light , $\lambda =2600\times {10}^{-10}m$

From Einstein's equation for photoelectric effect, $K{E}_{max}=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}$

or $K{E}_{max}=\frac{(6.62\times {10}^{-34})(3\times {10}^8)}{2600\times {10}^{-10}}-\frac{(6.62\times {10}^{-34})(3\times {10}^8)}{3800\times {10}^{-10}}=2.41\times {10}^{-19}J=1.51eV$ as $(1eV=1.6\times {10}^{-19}J)$