Question

Illuminating the surface of a certain metal alternately with light of wavelength ${\lambda }_2=0.54\mu m$ and ${\lambda }_2=0.54\mu m$, it was found that the corresponding maximum velocities of photo electrons have a ratio $\eta =2$, Find the work function of that metal.

• A. $6.99eV$
• B. $1.9eV$
• C. $7.53eV$
• D. $1.88eV$

Answer 1

The correct option is D $1.88eV$

$\eta =\frac{{\left(v_1\right)}_{max}}{{\left(v_2\right)}_{max}}=\sqrt{\frac{\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_o}}{\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_o}}}$

Thus, $n^2(\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_o})=\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_o}$

$⟹\frac{1}{{\lambda }_o}=\frac{(\frac{n^2}{{\lambda }_2}-\frac{1}{{\lambda }_1})}{(n^2-1)}$

So, $A=\frac{2\pi hc}{{\lambda }_o}=\frac{2\pi hc}{{\lambda }_2}\frac{(n^2-\frac{{\lambda }_2}{{\lambda }_1})}{(n^2-1)}=1.88eV$