Question

Protons of energy 6eV are indicated on a potassium surface of work function 2.1eV. What is the stopping potential?
A.−6V
B.−2.1V
C.−3.9V
D.−8.1V

Answer 1

Hint: Since the energy of the protons and the threshold energy of the potassium surface is given in the question, and we know the equation for the relationship between the energy of the photon and the kinetic energy of the emitted photoelectron is, \[{E_{photon}} = \varphi + {E_{electron}}\]. So, substituting the values in the equation, we can find the stopping potential

Complete step by step answer:
The relationship between the energy of the photon and the kinetic energy of the emitted photoelectron can be written as follows:
\[{E_{photon}} = \varphi + {E_{electron}}\]
\[ \Rightarrow hv = h{v_{th}} + \dfrac{1}{2}{m_e}{v^2}\]
Where,
\[{E_{photon}}\;\] denotes the energy of the incident photon, which is equal to h𝜈
φ denotes the threshold energy of the metal surface, which is equal to \[h{v_{th}}\]
\[{E_{electron}}\;\] denotes the kinetic energy of the photoelectron, which is equal to \[\dfrac{1}{2}{m_e}{v^2}\] ( \[{m_e}\; = \] mass of electron \[ = 9.1 \times {10^{ - 31}}\] kg)
Given in the question are,
\[hv = 6Ev\]
\[ \Rightarrow h{v_{th}}\; = 2.1Ev\]
Now substituting the values in the above equation,
\[ \Rightarrow hv = h{v_{th}} + \dfrac{1}{2}{m_e}{v^2}\]
\[ \Rightarrow \dfrac{1}{2}{m_e}{v^2} = {E_{electron}} = hv - h{v_{th}}\;\left( {6-2.1} \right) = 3.9V\]
Hence, the stopping potential is 3.9V.

Therefore, the correct answer is option (C).

Note: If the energy of the photon is less than the threshold energy, then there will not be emission of photoelectrons as the attractive forces between the nuclei and the electrons cannot be overcome. Thus, the photoelectric effect will not occur if \[v < {v_{th}}\]. If the frequency of the photon is exactly equal to the threshold frequency \[(v = {v_{th}})\], then there will be an emission of photoelectrons, but their kinetic energy will be equal to 0.