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A photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\frac{\lambda}{2}$. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is: (h = Planck's constant, c = speed of light)
A
$\frac{hc}{3\lambda}$
B
$\frac{hc}{2\lambda}$
C
$\frac{hc}{\lambda}$
D
$\frac{2hc}{\lambda}$
Detailed Solution
$KE_1 = \frac{hc}{\lambda} - \phi$; $KE_2 = \frac{2hc}{\lambda} - \phi$
$KE_2 = 3KE_1$: $\frac{2hc}{\lambda} - \phi = 3\left(\frac{hc}{\lambda} - \phi\right)$
$2\phi = \frac{hc}{\lambda} \Rightarrow \phi = \frac{hc}{2\lambda}$
$KE_2 = 3KE_1$: $\frac{2hc}{\lambda} - \phi = 3\left(\frac{hc}{\lambda} - \phi\right)$
$2\phi = \frac{hc}{\lambda} \Rightarrow \phi = \frac{hc}{2\lambda}$