Question

A photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda $ and $\dfrac{\lambda }{2}$. If the maximum kinetic energy of the emitted photoelectrons in the second case is three times that in the first case, the work function of the surface is:
$\begin{align}
  & a)\dfrac{hc}{2\lambda } \\
 & b)\dfrac{hc}{\lambda } \\
 & c)\dfrac{hc}{3\lambda } \\
 & d)\dfrac{3hc}{\lambda } \\
\end{align}$

Answer 1

Hint: In the above case it is given that a surface is illuminated with two different monochromatic light i.e.(single wavelength). The energy of the illuminated light is utilized to overcome the work function of the electrons of the particular surface and the remaining gets converted to the kinetic energy of the emitted electrons. Therefore we will consider this for both the monochromatic lights and accordingly obtain the work function of the material or the surface.
Formula used:
$E=h\gamma $
$h\gamma =W+K.{{E}_{MAX}}$
$c=\lambda \gamma $

Complete step-by-step solution:
The work function (W) is defined as the minimum amount of energy required by an electron to just escape from the metal surface. Let us say if we illuminate a surface with light from a monochromatic source with energy E, frequency $\gamma $ and wavelength $\lambda $, then some amount of energy will be utilized to overcome the work function of the surface, and the remaining will be seen in the form of the maximum kinetic energy of the emitted electrons. Mathematically this can be written as,
$\begin{align}
  & E=W+K.{{E}_{MAX}}\text{, }E=h\gamma \\
 & \Rightarrow h\gamma =W+K.{{E}_{MAX}}, \\
\end{align}$
where ‘h’ is Planck’s constant .Let us say the speed of light is equal to c then we can say that, $c=\lambda \gamma $.
Substituting this for frequency in the above equation we get,
$\begin{align}
  & h\gamma =W+K.{{E}_{MAX}}\text{, }\because \gamma =\dfrac{c}{\lambda } \\
 & \Rightarrow \dfrac{hc}{\lambda }=W+K.{{E}_{MAX}}...(1) \\
\end{align}$
Now let us consider that we illuminate the surface with a source of light whose wavelength is $\lambda $. Hence from the above equation 1 we get,
$\dfrac{hc}{\lambda }=W+K.{{E}_{MAX}}...(2)$
It is given in the question when the same surface is illuminated with wavelength of light i.e. $\dfrac{\lambda }{2}$ the kinetic energy is three times that in the first case. Hence from equation 1 we get,
$\begin{align}
  & \dfrac{hc}{\lambda /2}=W+3K.{{E}_{MAX}} \\
 & \Rightarrow \dfrac{2hc}{\lambda }=W+3K.{{E}_{MAX}}...(3) \\
\end{align}$
Now let us subtract equation 3 and 2 from which we get,
$\begin{align}
 & \dfrac{2hc}{\lambda }-\dfrac{hc}{\lambda }=W+3K.{{E}_{MAX}}-\left( W+K.{{E}_{MAX}} \right) \\
 & \Rightarrow \dfrac{hc}{\lambda }=W+3K.{{E}_{MAX}}-W-K.{{E}_{MAX}} \\
 & \Rightarrow \dfrac{hc}{\lambda }=3K.{{E}_{MAX}}-K.{{E}_{MAX}}=2K.{{E}_{MAX}} \\
 & \Rightarrow \dfrac{hc}{\lambda }=2K.{{E}_{MAX}} \\
 & \Rightarrow K.{{E}_{MAX}}=\dfrac{hc}{2\lambda } \\
\end{align}$
Now let us substitute the maximum kinetic energy obtained in equation 2 and we obtain,
$\begin{align}
  & \dfrac{hc}{\lambda }=W+K.{{E}_{MAX}} \\
 & \Rightarrow \dfrac{hc}{\lambda }=W+\dfrac{hc}{2\lambda } \\
 & \Rightarrow W=\dfrac{hc}{\lambda }-\dfrac{hc}{2\lambda }=\dfrac{hc}{2\lambda } \\
\end{align}$
Therefore the correct answer of the above question is option a.

Note: In the above scenario the surface on which light is emitted is kept the same. Therefore we have considered the work function to be also the same. The work function is different for different materials on which the light is illuminated.