AIIMS 2004 Physics Stopping Potential MCQ Question

To solve the problem regarding the photoelectric effect, we need to analyze the given information step by step.

Given Data

• Energy of the photon, $E = 4 \, \text{eV}$
• Work function of the metal, $\phi = 2 \, \text{eV}$

Step 1: Determine the Kinetic Energy of the Emitted Electrons

When a photon strikes a metal surface, it may transfer its energy to an electron. If the photon's energy exceeds the work function of the metal, electrons will be emitted. The energy of the emitted electrons is given by the equation:

$KE = E - \phi$

Substituting the given values:

$KE = 4 \, \text{eV} - 2 \, \text{eV} = 2 \, \text{eV}$

This means that each emitted electron has a kinetic energy of $2 \, \text{eV}$.

Step 2: Understanding Stopping Potential

The stopping potential ($V_s$) is the minimum reverse potential that must be applied to stop the emission of electrons. The stopping potential can be directly related to the kinetic energy of the emitted electrons. The relationship is given by:

$KE = e V_s$

where $e$ is the charge of an electron (in electron-volts, since we are using eV).

Step 3: Calculate the Stopping Potential

Since we already found that the kinetic energy of the emitted electrons is $2 \, \text{eV}$, we set this equal to $e V_s$:

$2 \, \text{eV} = e V_s$

Since $e$ in this context is 1 (because we are working in eV, where 1 eV corresponds to the energy gained by a charge of 1 Coulomb moving through a potential of 1 volt), we can rearrange this equation to find $V_s$:

$V_s = 2 \, \text{V}$

Conclusion

Thus, the minimum reverse potential that needs to be applied to stop the emission of electrons is 2 V.

Correct Answer

The correct answer is A) 2 V.

Why Other Options Are Incorrect

• B) 4 V: This would suggest the kinetic energy of the emitted electrons is 4 eV, which is not the case. The kinetic energy calculated was only 2 eV.
• C) 6 V: Similar reasoning applies here; it does not relate to the kinetic energy found from the photon and work function.
• D) 8 V: This option is even further away from the calculated kinetic energy and is incorrect for the same reason.

In summary, the stopping potential is directly derived from the kinetic energy of the emitted electrons, which was determined to be 2 eV, leading us to the conclusion that the stopping potential must also be 2 V.