Question

It has been found that, when the surface of metal is irradiated with radiation of suitable frequency, electrons (photo electrons) are ejected from the surface of metal. Minimum energy required for ejection of electron from metal surface is called work function of metal.

A metallic surface is irradiated alternatively with radiations of wavelength 3000$A^{\circ }$ and 6000$A^{\circ }$. It is observed that the maximum speeds of the photoelectrons under these cases are in the ratio 3:1.
Work function of the metallic surface is?

• A. 3.25 eV
• B. 1.81 eV
• C. 9.12 eV
• D. 5.21 eV

Answer 1

Answer: (B)

1.81 eV

$(KE{)}_{max}=\frac{hc}{\lambda }-{\varphi }_{\circ }$

where ${\varphi }_{\circ }=$ work function,

$\frac{1}{2}m{\left(3V\right)}^2=\frac{hc\times {10}^{10}}{3000}-{\varphi }_{\circ }$

$\frac{1}{2}m{V}^2=\frac{hc\times {10}^{10}}{6000}-{\varphi }_{\circ }$

$\therefore 9(\frac{hc\times {10}^{10}}{6000}-{\varphi }_{\circ })=\frac{hc\times {10}^{10}}{3000}-{\varphi }_{\circ }$

$\frac{9hc\times {10}^7}{6}-\frac{hc\times {10}^7}{3}=8\varphi$

${\varphi }_{\circ }=\frac{1}{8}\left(\frac{9-2}{6}\right)\times hc\times {10}^7=2.89\times {10}^{-19}J$

$=\frac{2.89\times {10}^{-19}}{1.6\times {10}^{-19}}eV=1.8eV$