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**Hint:**In the given question, we have to analyse the photoelectric effect from a metal surface. We have been given the work function of the metal surface and we are asked to find out the maximum wavelength that initiates photoelectric emissions. We all know that the maximum wavelength corresponds to the minimum energy of the photon and the minimum energy photon that can initiate photoelectric emission has an energy equal to the work function of the metal. Similarly, once we find the maximum kinetic energy of photoelectrons emitted, we can use that energy to find the retarding potential. Let’s see the detailed solution given below.

Formula Used: \[E=\dfrac{hc}{\lambda }\] , \[K=E-\phi \] , \[V=\dfrac{K}{e}\]

**Complete step by step solution:**

As discussed above, the maximum wavelength of the photon which can eject electrons from the metal will have energy equal to the work function of the metal.

We know that the relation between the energy of the photon, the speed of electromagnetic radiation, the Planck’s constant, and the wavelength of the photon is given as \[E=\dfrac{hc}{\lambda }\] where \[h\] is the Planck’s constant, \[c\] is the speed of light and, \[\lambda \] is the wavelength of the photon

In the case of maximum wavelength, the energy is the same as the work function, so the equation now becomes \[\phi =\dfrac{hc}{\lambda }\]

Work function, $\phi = 2.4eV = 2.4 \times 1.6 \times {10^{ - 19}}J(\because 1eV = 1.6 \times {10^{ - 19}}J)$

Rearranging the quantities and substituting the values, we get

\[\begin{align}

& \lambda =\dfrac{hc}{\phi } \\

& \Rightarrow \lambda =\dfrac{6.62\times {{10}^{-34}}\times 3\times {{10}^{8}}}{2.4\times 1.6\times {{10}^{-19}}}\left[ \because \left( h \right)=6.62\times {{10}^{-34}},\left( c \right)=3\times {{10}^{8}} \right] \\

& \Rightarrow \lambda =5.17\times {{10}^{-7}}m \\

\end{align}\]

The relation between the maximum kinetic energy of the emitted photon, the energy of the incident photon, and the work function can be given as \[K=E-\phi \] where \[K\] is the maximum kinetic energy of the emitted photon, \[E\] is the energy of the incident photon and \[\phi \] is the work function of the metal

Energy of the incident photon, $E = 3eV$

Substituting the values in the above equation, we get

\[\begin{align}

& K=E-\phi \\

& \Rightarrow K=\left( 3-2.4 \right)eV \\

& \Rightarrow K=0.6eV \\

\end{align}\]

The potential that stops the emitted photon from having maximum kinetic energy is known as the retarding potential.

The retarding potential of the metal can be given as \[(V)=\dfrac{K}{e}\] where \[K\] is the maximum kinetic energy of the emitted photon and \[e\] is the elementary charge

Substituting the values, we get

\[\begin{align}

& V=\dfrac{K}{e} \\

& \Rightarrow V=\dfrac{0.6eV}{e}=0.6V \\

\end{align}\]

**Note:**

For reaching the correct answer, we should be aware of the basics of the photoelectric effect. When the photon is incident on a metal surface, an energy equal to the work function of the metal is lost and the remaining energy is manifested in the emitted photon as the kinetic energy. Students should also know that division of energy in electron-volts with the elementary charge gives the voltage in volts.

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