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Question: If the wavelength of the incident radiation changes from \[{{\lambda }_{1}}\]to \[{{\lambda }_{2}}\]...

If the wavelength of the incident radiation changes from λ1{{\lambda }_{1}}to λ2{{\lambda }_{2}}, then the maximum kinetic energy of the emitted photoelectrons change from K1{{K}_{1}} to K2{{K}_{2}}, then the work function of the emitter surface is

A. λ1K1−λ2K2λ2−λ1A.\,\dfrac{{{\lambda }_{1}}{{K}_{1}}-{{\lambda }_{2}}{{K}_{2}}}{{{\lambda }_{2}}-{{\lambda }_{1}}}
B. λ1K2−λ2K1λ1−λ2B.\,\dfrac{{{\lambda }_{1}}{{K}_{2}}-{{\lambda }_{2}}{{K}_{1}}}{{{\lambda }_{1}}-{{\lambda }_{2}}}
C. K2−K1λ1K1−λ2K2C.\,\dfrac{{{K}_{2}}-{{K}_{1}}}{{{\lambda }_{1}}{{K}_{1}}-{{\lambda }_{2}}{{K}_{2}}}
D. λ2−λ1λ2K1−λ1K2D.\,\dfrac{{{\lambda }_{2}}-{{\lambda }_{1}}}{{{\lambda }_{2}}{{K}_{1}}-{{\lambda }_{1}}{{K}_{2}}}

Explanation

Solution

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:
Kmax⁡=hf−Φ{{K}_{\max }}=hf-\Phi
Where hfhf is the incident photon energy and Φ\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,
Kmax⁡=hf−Φ{{K}_{\max }}=hf-\Phi
Where hfhfis the incident photon energy and Φ\Phi is the work function.
hf=hcλhf=\dfrac{hc}{\lambda }
So, we get,
Kmax⁡=hcλ−Φ{{K}_{\max }}=\dfrac{hc}{\lambda }-\Phi

Now consider,
For the incident wavelength of λ1{{\lambda }_{1}}, the maximum kinetic energy of the emitted photoelectrons is K1{{K}_{1}}.
Thus, the equation is,
K1=hcλ1−Φ{{K}_{1}}=\dfrac{hc}{{{\lambda }_{1}}}-\Phi
hc=K1λ1+λ1Φhc={{K}_{1}}{{\lambda }_{1}}+{{\lambda }_{1}}\Phi …… (1)

For the incident wavelength of λ2{{\lambda }_{2}}, the maximum kinetic energy of the emitted photoelectrons is K2{{K}_{2}}.
Thus, the equation is,
K2=hcλ2−Φ{{K}_{2}}=\dfrac{hc}{{{\lambda }_{2}}}-\Phi
hc=K2λ2+λ2Φhc={{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,

& {{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{aligned}$$ Rearrange the terms to obtain the expression for the work function. Thus, $$\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{aligned} & 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{aligned}$$ Where $${{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.