Question

When a point source of light is at a distance of one metre from a photocell, the cut off voltage is found to be. V. If the same source is placed at 2 m distance from photocell, the cut of voltage will be:
A. \[V\]
B. \[V/2\]
C. \[V/4\]
D. \[V/\sqrt 2 \]

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 calculated as 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. This stopping potential does not depend upon the distance between light source and metal surface because that only varies the intensity of the light but not frequency or wavelength and thus also does not affect the kinetic energy of the electrons.

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 1 m away from a photocell, stopping the potential V is obtained In this photoelectric experiment. The distance is increased to 2 m. we need to find a new stopping potential.

Equation for photoelectric effect is,
\[h\nu = {\phi _o} + {E_k}\]---- (1)
\[\Rightarrow {\phi _o} = \dfrac{{hc}}{{{\lambda _o}}} \\ \]----(2)
\[\Rightarrow {E_k} = e{V_o}\]---- (3)
Here, \[{E_k}\] 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 V

Hence option A is the correct answer.

Additional information: Einstein developed the photoelectric effect in 1905, for which he received the Nobel Prize in Physics. According to its hypothesis, when sufficient energy light is incident on a metal surface, the photons of the incident light transmit energy to the electrons on the metallic surface. If this energy is greater than the metal's threshold energy, the electrons become sufficiently energetic to exit the metal surface and are expelled. These electrons are known as photoelectrons because their energy is less than that of the incoming light since part of it is used to overcome the barrier energy or the work function.

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 \[{E_k} = e{V_o}\].