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# Stopping Potential – Definition, Practice Problems and FAQ

Most of us would be aware of the story of Sisyphus. In Greek mythology, The creator and ruler of Ephyra in Greek mythology was Sisyphus, also known as Sisyphos. Zeus made him roll a huge rock up a hill, only for it to tumble down every time it got close to the summit, repeating this deed for all of eternity as a punishment for cheating death twice.

Why are we discussing ancient Greek mythology on this concept page? Does it have any relevance here?

It does have relevance. The minimum energy required by Sisyphus to stop the stone from rolling back down is the stopping potential or stopping energy of the rock. Similarly, in the photoelectric effect, when light of suitable frequency is incident on a metal surface, electrons are ejected from it. The energy or potential required to stop the ejection of electrons is called the stopping potential.

Let’s get to know more about stopping potential on this concept page!

• Stopping Potential
• Practice Problems
• Frequently Asked Questions – FAQ

## Stopping Potential

Stopping potential is the potential required to stop the removal of electrons from the surface of a metal when an incident light with energy greater than the work function of the metal is incident on it. With regards to the photoelectric effect, stopping potential and work function means the same thing.

When light strikes a metal surface, a current starts to flow. This current starts to drop if a retarding potential is supplied to prevent the electrons from travelling to the other metal surface. When the retarding potential is applied, the electrons would not have enough kinetic energy to pass the potential barrier at some point, causing the current to cease. The stopping potential (cut-off potential) is the potential at which the current equals zero. The stopping potential refers to the maximum amount of voltage that can be measured in a lab.

The maximum kinetic energy of the expelled electrons can be calculated by knowing the stopping (cut-off) voltage. The stopping potential, which is proportional to the kinetic energy of the photoelectron, is the lowest potential necessary to stop the photoelectric current.

• The amount of incident light does not affect stopping potential. The value of saturated current rises with intensity while the stopping potential stays constant. The amount of incident radiation does not affect the stopping potential.
• The stopping potential is frequency-dependent for a given radiation intensity. The stopping potential increases with increasing light frequency.

⇒ If stopping potential = Vs, then K.E.max = ${e×V}_{s}$

Where ‘e’ is the charge of an electron having a value of $1.6×{10}^{-19}C$

## Practice Problems

Q1. Small light packs are referred to as

A. Photon
B. Quanta
C. Proton
D. All of these

Solution: Quantum of light refers to the smallest amount of energy that may be either emitted or absorbed in the form of electromagnetic radiation. The light energy emanating from any source, according to Planck, is always an integral multiple of the smallest energy value known as the quantum of light.

If the energy contained in a single photon of light is equal to E0(J), then the total energy leaving the object is equal to nE0(J) (n = Integer).

Photon = Quantum of Light (Packet or bundle of energy)

Energy of one photon is given by E0=hv (v- Frequency of light) h = $6.626×{10}^{-34}Js$(h - Planck constant)

So, option A is the correct answer.

Q2. A radiation of 4000 Å falls on a metal surface for which the work function is 1.25 eV. What is the stopping potential of ejected electrons?

A. 2.81 V
B. 1.76 V
C. 1.21 V
D. 4.496 V

Solution: We know that, Kinetic energy =

Kinetic energy = …………… (1)

We know that, Kinetic energy =eVs …………… (2)

Substituting equation (2) in (1),

eVs= …………… (3)

Where,

Wavelength of radiation = 4000 Å =

Substituting the values in equation (3),

Vs=1.76 V

So, option B is the correct answer.

Q3. When electromagnetic radiation of wavelength 413.5 nm falls on the surface of caesium, electrons are emitted with kinetic energy =1.75 eV. Determine the work function (W0) of cesium.

A. 1.25 eV
B. 1.75 eV
C. 2 eV
D. 3.65 eV

Solution: We know that

So, option A is the correct answer.

Q4.  A beam of wavelength 178 nm falls on the surface of a metal. The resulting stopping voltage is 2.32 V. Calculate the work function (in eV) of the metal.

A. 3.24 V
B. 2.56 V
C. 7.98 V
D. 4.65 V

Solution: Energy of the incident photon =

Stopping voltage =2.32 V

Kinetic energy of the electron =2.32 V

Work function

So, option D is the correct answer.

## Frequently Asked Questions – FAQ

Q1.  If we compare Li, Pt and Cs, which has the lowest work function?
Cs (Caesium) is the most electropositive metal in the periodic table. Therefore, it has the lowest value of the work function. It is directly related to ionization energy.

Q2. What are the factors affecting stopping potential?