The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential in volts is :

1. Break Down the Problem

To solve this problem, we need to determine the stopping potential $V_s$. The stopping potential is the potential needed to stop the fastest photoelectrons emitted from a surface when illuminated by light.

2. Relevant Concepts

The maximum kinetic energy $KE_{\text{max}}$ of the photoelectrons is related to the energy of the incoming photons $E_{photon}$ by the equation: $KE_{\text{max}} = E_{photon} - \Phi$ where $\Phi$ is the work function of the material. The stopping potential $V_s$ can also be determined by the equation relating kinetic energy to potential: $KE_{\text{max}} = e \cdot V_s$ where $e$ is the charge of an electron (approximately $1.6 \times 10^{-19}$ coulombs).

3. Analysis and Detail

From the problem, we know:

• The energy of the photons, $E_{photon} = 6$ eV
• Maximum kinetic energy, $KE_{\text{max}} = 4$ eV

Using the relation for kinetic energy, we can determine the work function: $4 \, \text{eV} = 6 \, \text{eV} - \Phi$ Rearranging gives us: $\Phi = 6 \, \text{eV} - 4 \, \text{eV} = 2 \, \text{eV}$

Now, using the relation for stopping potential with $KE_{\text{max}} = e \cdot V_s$: $4 \, \text{eV} = e \cdot V_s$ Thus, $V_s$ is calculated directly as: $V_s = \frac{KE_{\text{max}}}{e} = 4 \, \text{V}$

4. Verify and Summarize

We have derived $V_s$ using the kinetic energy and the concept of stopping potential. The calculations are consistent and verified through the stopping potential relationship.

Final Answer

The stopping potential in volts is $\boxed{4 \, \text{V}}$.