Question

A photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$. 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. $\frac{2hc}{\lambda }$
• B. $\frac{hc}{3\lambda }$
• C. $\frac{hc}{2\lambda }$
• D. $\frac{hc}{\lambda }$

Answer 1

The correct option is C $\frac{hc}{2\lambda }$

Let ${\varphi }_0$ be the work function of the surface of material. Then,
$K_{max1}=\frac{hc}{\lambda }-{\varphi }_0\mathrm{in}\mathrm{first}\mathrm{case}\mathrm{and}{K}_{max2}=\frac{hc}{\frac{\lambda }{2}}-{\varphi }_0=\frac{2hc}{\lambda }-{\varphi }_0\mathrm{in}\mathrm{second}\mathrm{case}\mathrm{But},K_{max2}=3K_{max1}\left(given\right)\therefore \frac{2hc}{\lambda }-{\varphi }_0=3\left(\frac{hc}{\lambda }-{\varphi }_0\right)\Rightarrow 3{\varphi }_0-{\varphi }_0=\frac{3hc}{\lambda }-\frac{2hc}{\lambda }\Rightarrow 2{\varphi }_0=\frac{hc}{\lambda }\Rightarrow {\varphi }_0=\frac{hc}{2\lambda }$