Question

If the wavelength of the incident radiation changes from \[{{\lambda }_{1}}\]to \[{{\lambda }_{2}}\], then the maximum kinetic energy of the emitted photoelectrons change from \[{{K}_{1}}\] to \[{{K}_{2}}\], then the work function of the emitter surface is

\[A.\,\dfrac{{{\lambda }_{1}}{{K}_{1}}-{{\lambda }_{2}}{{K}_{2}}}{{{\lambda }_{2}}-{{\lambda }_{1}}}\]
\[B.\,\dfrac{{{\lambda }_{1}}{{K}_{2}}-{{\lambda }_{2}}{{K}_{1}}}{{{\lambda }_{1}}-{{\lambda }_{2}}}\]
\[C.\,\dfrac{{{K}_{2}}-{{K}_{1}}}{{{\lambda }_{1}}{{K}_{1}}-{{\lambda }_{2}}{{K}_{2}}}\]
\[D.\,\dfrac{{{\lambda }_{2}}-{{\lambda }_{1}}}{{{\lambda }_{2}}{{K}_{1}}-{{\lambda }_{1}}{{K}_{2}}}\]

Answer 1

Hint: As the maximum kinetic energy is the difference between the incident photon energy and the work function, the given problem can be simplified by considering 2 situations and then resolving further considering the same work function for both the situations.

Formula used:
\[{{K}_{\max }}=hf-\Phi \]
Where \[hf\] is the incident photon energy and \[\Phi \] is the work function.

Complete answer:
The maximum kinetic energy of the emitted photoelectrons is equal to the difference between the incident photon energy and the work function.

The maximum kinetic energy of the emitted photoelectrons is given by,
\[{{K}_{\max }}=hf-\Phi \]
Where \[hf\]is the incident photon energy and \[\Phi \] is the work function.
\[hf=\dfrac{hc}{\lambda }\]
So, we get,
\[{{K}_{\max }}=\dfrac{hc}{\lambda }-\Phi \]

Now consider,
For the incident wavelength of \[{{\lambda }_{1}}\], the maximum kinetic energy of the emitted photoelectrons is \[{{K}_{1}}\].
Thus, the equation is,
\[{{K}_{1}}=\dfrac{hc}{{{\lambda }_{1}}}-\Phi \]
\[hc={{K}_{1}}{{\lambda }_{1}}+{{\lambda }_{1}}\Phi \]…… (1)

For the incident wavelength of \[{{\lambda }_{2}}\], the maximum kinetic energy of the emitted photoelectrons is \[{{K}_{2}}\].
Thus, the equation is,
\[{{K}_{2}}=\dfrac{hc}{{{\lambda }_{2}}}-\Phi \]
\[hc={{K}_{2}}{{\lambda }_{2}}+{{\lambda }_{2}}\Phi \]…… (2)

As the LHS parts of the equations (1) and (2) are same, thus, equate the equations (1) and (2).
So, we get,
\[\begin{align}
  & {{K}_{1}}{{\lambda }_{1}}+{{\lambda }_{1}}\Phi ={{K}_{2}}{{\lambda }_{2}}+{{\lambda }_{2}}\Phi \\
 & {{K}_{1}}{{\lambda }_{1}}-{{K}_{2}}{{\lambda }_{2}}={{\lambda }_{2}}\Phi -{{\lambda }_{1}}\Phi \\
 & {{K}_{1}}{{\lambda }_{1}}-{{K}_{2}}{{\lambda }_{2}}=\Phi ({{\lambda }_{2}}-{{\lambda }_{1}}) \\
\end{align}\]

Rearrange the terms to obtain the expression for the work function.
Thus,
\[\Phi =\dfrac{{{K}_{1}}{{\lambda }_{1}}-{{K}_{2}}{{\lambda }_{2}}}{{{\lambda }_{2}}-{{\lambda }_{1}}}\]

So, the correct answer is “Option A”.

Note:
 The things to be on your figure tips for further information on solving these types of problems are:
The work function is denoted by \[\Phi \] and even \[W\].
There are 3 equations for the work function.
\[\begin{align}
  & E={{W}_{0}}+K{{E}_{\max }} \\
 & hf={{W}_{0}}+\dfrac{1}{2}m{{v}_{\max }}^{2} \\
 & {{W}_{0}}=h{{f}_{0}}=\dfrac{hc}{{{\lambda }_{0}}} \\
\end{align}\]
Where \[{{W}_{0}}\] is the work function, \[{{f}_{0}}\] is the threshold frequency and \[{{\lambda }_{0}}\] is the threshold wavelength.
Use the formulae according to the given situations.