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1800-102-2727Most 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!
TABLE OF CONTENTS
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.
⇒ If stopping potential = V_{s}, then K.E._{max} = ${e\times V}_{s}$
Where ‘e’ is the charge of an electron having a value of $1.6\times {10}^{-19}C$
Q1. Small light packs are referred to as
A. Photon
B. Quanta
C. Proton
D. All of these
Answer: A
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 E_{0}(J), then the total energy leaving the object is equal to nE_{0}(J) (n = Integer).
Photon = Quantum of Light (Packet or bundle of energy)
Energy of one photon is given by E_{0}=hv (v- Frequency of light) h = $6.626\times {10}^{-34}Js$(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
Answer: B
Solution: We know that, Kinetic energy = $h\upsilon -h{\upsilon}_{0}=h\upsilon -\varphi $
Kinetic energy = $\frac{hc}{\lambda}-\varphi $ …………… (1)
We know that, Kinetic energy =eV_{s} …………… (2)
Substituting equation (2) in (1),
eV_{s}= $\frac{hc}{\lambda}-\varphi $ …………… (3)
Where, $e=chargeofelectron\left(1.6\times {10}^{-19}C\right)$
${V}_{s}=stoppingvoltage$
$c=speedoflight\left(3\times {10}^{8}m{s}^{-1}\right)$
Wavelength of radiation = 4000 Å = $4000\times {10}^{-10}m$
Substituting the values in equation (3),
$1.6\times {10}^{-19}C\times {V}_{s}=\frac{6.626\times {10}^{-34}Js\times 3\times {10}^{8}m{s}^{-1}}{4000\times {10}^{-10}m}-1.35\times 1.6\times {10}^{-19}C$
$1.6\times {10}^{-19}C\times {V}_{s}=4.97\times {10}^{-19}-2.16\times {10}^{-19}$
V_{s}=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 (W_{0}) of cesium.
A. 1.25 eV
B. 1.75 eV
C. 2 eV
D. 3.65 eV
Answer: A
Solution: We know that
$E=\frac{hc}{{\lambda}_{}}=\frac{1240}{{\lambda}_{nm}}eV(when\lambda isinnm)\phantom{\rule{0ex}{0ex}}E=\frac{hc}{{\lambda}_{}}=\frac{12400}{{\lambda}_{{A}^{0}}}eV(when\lambda isin{A}^{0})\phantom{\rule{0ex}{0ex}}E=\frac{1240}{413.5}eV\approx 3eV\phantom{\rule{0ex}{0ex}}{\mathit{W}}_{0}=\frac{hc}{{\lambda}_{}}-KE\phantom{\rule{0ex}{0ex}}=3eV-1.75eV=1.25eV\phantom{\rule{0ex}{0ex}}$
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
Answer: D
Solution: Energy of the incident photon = $\frac{1240}{178nm}eV=6.97eV$
Stopping voltage =2.32 V
Kinetic energy of the electron =2.32 V
Work function $\left(\varphi \right)=6.97-2.32eV=4.65eV$
So, option D is the correct answer.
Q1. If we compare Li, Pt and Cs, which has the lowest work function?
Answer: 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?
Answer: The frequency of the incident light, not its intensity, determines the stopping potential. The kinetic energy of the electrons, which is dependent on the stopping potential, is only changed by the frequency of the incoming light and not by its intensity.
Q3. What is the relationship between wavelength and stopping potential?
Answer: Higher stopping potentials are produced by light with shorter wavelengths. This demonstrates that shorter wavelengths and higher energy frequencies have more energy to transmit to the photoelectrons. As a result, the photoelectrons receive more energy from higher frequencies and are able to exit the metal with more kinetic energy.
Q4. Why does the number of ejected electrons is increased with an increase in the intensity of light?
Answer: The amount of photoelectrons released is influenced by the intensity of the incident light. The number of incident photons and, consequently, the number of electrons emitted from the metal surface is directly proportional to the intensity of the incident light. Therefore, as the intensity of the incident radiation increases, more electrons will be expelled.