Question

When the light source is kept 20cm away from a photocell, stopping the potential of 0.6V is obtained. When the source is kept 40cm away, the stopping potential will be:
A. \[0.3V\]
B. \[0.6V\]
C. \[1.2V\]
D. \[2.4V\]

Answer 1

Hint:The equation for photoelectric effect is \[h\nu = {\phi _o} + {E_k}\]. The minimum amount of energy required for electron emission from the metal surface is the work function of that metal and the frequency of light corresponding to this minimum energy is called threshold frequency and the corresponding wavelength is called threshold wavelength. It is calculated as a work function, \[{\phi _o} = h{\nu _o} = \dfrac{{hc}}{{{\lambda _0}}}\]. \[{E_k}\] is the energy of the emitted electron. It is also known as the maximum kinetic energy such that \[{E_k} = e{V_o}\]where \[{V_o}\] is the stopping potential

Formula(e) Used:
Energy, \[E = h\nu = \dfrac{{hc}}{\lambda }\]
Equation for photoelectric effect, \[h\nu = {\phi _o} + {E_k}\]
Work function, \[{\phi _o} = h{\nu _o} = \dfrac{{hc}}{{{\lambda _o}}}\]
Maximum kinetic energy, \[{E_k} = e{V_o}\]
Where
E = energy of incident radiation
\[{\phi _o}\]= Work function of the metal
\[{E_k}\]= energy of the emitted photoelectron
\[c{\rm{ }} =\text{speed of light} = 3 \times {10^8}m/s\]
\[\nu \]= frequency of the light,
\[{\nu _o}\]= threshold frequency,
\[\lambda \]= wavelength of the light,
\[{\lambda _o}\]= threshold wavelength
\[e = 1.6 \times {10^{ - 19}}C\]= electronic charge
\[{V_o}\]= stopping potential

Complete step by step solution:
Given: When the light source is kept 20 cm away from a photocell, stopping the potential of 0.6V is obtained In this photoelectric experiment. The distance is increased to 40 cm. We need to find a new stopping potential.
\[c{\rm{ }} = \text{speed of light} = 3 \times {10^8}m/s\]

Photoelectric effect was discovered by Einstein in 1905 for which he also won the Nobel prize in physics. According to its theory, when a light of sufficient energy is incident on a metal surface, the photons of the incident light impart energy to the electrons on the metallic surface. If this energy is higher than the threshold energy of the metal, the electrons become sufficiently energetic to escape the metal surface and are emitted. These electrons are called photoelectrons whose energy is less than the energy of the incident light as some of the energy is utilised in overcoming the barrier energy or the work function.

Equation for photoelectric effect is,
\[h\nu = {\phi _o} + {E_k}\]---- (1)
\[{\phi _o} = \dfrac{{hc}}{{{\lambda _o}}}\]----(2)
\[{E_k} = e{V_o}\]---- (3)
\[{E_k}\] is also the maximum kinetic energy

When the distance between the light source and the photocell is increased, the intensity of the incident radiation falling on the metallic surface decreases. However, the wavelength and frequency of the incident light remains unchanged. So, the maximum kinetic energy will remain unchanged as it is directly related to the stopping potential which depends on the change in frequency and wavelength.

The work function is the property of the material and independent of the intensity of the incident light. So, the stopping potential will not change with increasing distance of the source from the photocell. Thus, the new stopping potential is 0.6 V

Hence option B is the correct answer.

Note: Work function is a property of the metal that depends on the metal. The remaining energy after the work function is responsible for emission of the electron from the surface. This extra energy is converted to kinetic energy which enables the electron to emit from the metal surface. If we know the energy of the incident radiation and the threshold, we can determine the additional kinetic energy and the stopping potential using \[{E_k} = e{V_o}\].