Question

Surface of certain metal is first illuminated with light of wavelength $\lambda_1 =350 \; nm$ and then, by light of wavelength $\lambda_2=54\, \; nm$. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of $2$. The work function of the metal (in eV) is close to : (Energjr of photon = $\frac{1240}{\lambda (in \; nm)} eV$)

• A. 1.8
• B. 1.4
• C. 2.5
• D. 5.6

Answer 1

The Correct Option is A

$\frac{hc}{\lambda_{1}} =\phi + \frac{1}{2} m\left(2v\right)^{2} $ $\frac{hc}{\lambda _{2}} =\phi + \frac{1}{2} mv^{2} $ $\Rightarrow \frac{\frac{hc}{\lambda _{1}}-\phi}{\frac{hc}{\lambda _{2}} -\phi} = 4 $ $\Rightarrow \frac{hc}{\lambda _{1}} - \phi = \frac{4hc}{\lambda _{2}} -4 \phi $ $ \Rightarrow \frac{4hc}{\lambda _{2}} - \frac{hc}{\lambda _{1}} = 3\phi $ $ \Rightarrow \phi = \frac{1}{3} hc \left(\frac{4}{\lambda _{2}} -\frac{1}{\lambda _{1}}\right) $ $= \frac{1}{3} \times1240 \left(\frac{4 \times350 -540}{350\times540} \right) $ $= 1.8 \,eV $