Question

A sodium metal piece is illuminated with light of wavelength $0.3\mu m$. The work function of sodium is 2.46 eV. For this situation, mark out the correct statement(s)

• A. The maximum kinetic energy of the ejected photoelectrons is 1.67 eV
• B. The cut-off wavelength for sodium is 505 nm
• C. The minimum photon energy of incident light for photoelectric effect to take place is 2.46 eV
• D. All of the above

Answer 1

The correct option is D All of the above

The maximum kinetic energy of photoelectrons ,

$K{E}_{max}=\frac{hc}{\lambda }-W=\frac{(6.62\times {10}^{-34})(3\times {10}^8)}{(0.3\times {10}^{-6})(1.6\times {10}^{-19})}eV-2.46eV=1.67eV$

If ${\lambda }_0$ is the cut-off wavelength, $W=\frac{hc}{{\lambda }_0}$

or ${\lambda }_0=\frac{hc}{W}=\frac{(6.62\times {10}^{-34})(3\times {10}^8)}{2.46\times 1.6\times {10}^{-19}}=5.05\times {10}^{-7}m=505\times {10}^{-9}=505nm$

We know that the work function is the minimum photon energy for taking place of photoelectric effect.