A photocell stops emission if it is maintained at 2V negative potential. Find the energy of the most energetic photoelectron.
A. 2eV
B. 2J
C. 2KJ
D. 2KeV

Hint:Before we proceed with the given problem. Let us understand the stopping potential. Stopping potential is the minimum negative voltage that is applied to the anode to stop the photocurrent. The maximum kinetic energy of the electrons will be equal to the stopping voltage when it is measured in electron volt.

Formula Used:
The formula to find the kinetic energy of an electron is,
\[K.{E_{\max }} = {V_0}\left( {eV} \right)\]
Where, \[{V_0}\] is stopping potential.

Complete step by step solution:
Consider a photocell that stops emitting photoelectrons if it is maintained at 2V negative potential. That is \[{V_0} = - 2V\]. Now we need to find the energy of the most energetic photoelectron. In order to find the energy of the energetic photoelectron, that is nothing but the kinetic energy of the photoelectron which is given as,
\[K.{E_{\max }} = {V_0}\left( {eV} \right)\]
By taking the magnitude of stopping potential we can write the above equation as,
\[K.{E_{\max }} = \left| {{V_0}} \right|\left( {eV} \right)\]
Substitute the value of \[{V_0} = - 2V\] in above equation we get,
\[K.{E_{\max }} = 2\,eV\]
Therefore, the energy of the most energetic photoelectron is 2 eV.

Hence, Option A is the correct answer.

Note: Remember that the kinetic energy carried by the photoelectron cannot be negative. Therefore, while calculating the kinetic energy, that is, \[K.{E_{\max }} = {V_0}\left( {eV} \right)\]. The stopping potential given in eV is negative, so we have to take only the magnitude of the stopping potential and then we calculate the kinetic energy of an electron as discussed in this problem.