Source: https://hemantmore.org.in/science/physics/stopping-potential/12631/
Timestamp: 2019-04-20 05:08:57+00:00

Document:
When a radiation of certain wavelength shines on the cathode of the photoelectric cell, the photocurrent produced can be reduced to zero by applying stopping potential of 2.63 V. If the work function of the photoemitter is 4 eV, find the wavelengths of radiation.
Given: Stopping potential = Vs = 2.63 V, work function = Φ = 4 eV = 4 x 1.6 x 10-19 J = 6.4 x 10-19 J, speed of light = c = 3 x 108 m/s, Planck’s constant = h = 6.63 x 10-34 Js, Charge on electron = e = 1.6 x 10-19 C.
To Find: Wavelength of radiation = λ = ?
Radiation of wavelength 3000 A falls on a photoelectric surface for which work function is 1.6 eV. What is the stopping potential for emitted electron.
Given: Wavelength of radiation = λ = 3000 Å = 3000 x 10-10 m, work function = Φ = 1.6 eV = 1.6 x 1.6 x 10-19 J = 2.56 x 10-19 J, speed of light = c = 3 x 108 m/s, Planck’s constant = h = 6.63 x 10-34 Js, Charge on electron = e = 1.6 x 10-19 C.
To Find: Stopping potential = Vs =?
Ans: The stopping potential is 2.54 V.
When a radiation of certain wavelength shines on the cathode of the photoelectric cell, the photocurrent produced can be reduced to zero by applying stopping potential of 3 V. If the work function of the photoemitter is 3.63 eV, find the frequency of radiation.
Given: Stopping potential = Vs = 3 V, work function = Φ = 3.63 eV = 3.63 x 1.6 x 10-19 J, speed of light = c = 3 x 108 m/s, Planck’s constant = h = 6.63 x 10-34 Js, Charge on electron = e = 1.6 x 10-19 C.
To Find: frequency of radiation = ν = ?
Photoelectrons emitted by a surface have maximum kinetic energy of 4 x 10-19 J. What is the stopping potential for phpto emission from the surface for the incident radiation.
Given: Maximum kinetic energy of photoelectron = K.E. max = 4 x 10-19 J, Charge on electron = e = 1.6 x 10-19 C.
Light of wavelength 2000 Å is incident on the cathode of a photocell. The current in the photocell is reduced to zero by stopping potential of 2 V. Find the threshold wavelength of the material of cathode.
Given: Stopping potential = Vs = 2 V, wavelength of incident light = λ = 2000 Å = 2000 x 10-10 m, speed of light = c = 3 x 108 m/s, Planck’s constant = h = 6.63 x 10-34 Js, Charge on electron = e = 1.6 x 10-19 C.
Photoelectrons are ejected fro metal surface when radiation of wavelength 160 nm is incident on the surface. Find stopping potential of emitted electrons if the limiting wavelength is 240 nm for photoelectric emission from the surface.
Given: Stopping potential = Vs = 2 V, wavelength of incident light = λ = 160 nm = 160 x 10-9 m, Threshold wavelength = λo = 240 nm = 240 x 10-9 m, , speed of light = c = 3 x 108 m/s, Planck’s constant = h = 6.63 x 10-34 Js, Charge on electron = e = 1.6 x 10-19 C.
To Find:Stopping potential = Vs =?
Calculate the change in stopping potential when the wavelength of light incident on photoelectric surface is reduced from 4000 Å to 3600 Å.
Given: Initial wavelength = λ1 = 4000 Å = 4000 x 10-10 m = 4 x 10-7 m, Final wavelength = λ2 = 3600 Å = 3600 x 10-10 m = 3.6 x 10-7 m, speed of light = c = 3 x 108 m/s, Planck’s constant = h = 6.63 x 10-34 Js, Charge on electron = e = 1.6 x 10-19 C.
Given: Initial frequency = ν1 = 2.2 x 1015 Hz, initial stopping potential = Vs1 =6.6 V, Final frequency = ν2 = 4.6 x 1015 Hz, Final stopping potential = Vs2 = 16.5 V, Charge on electron = e = 1.6 x 10-19 C.
To Find: Planck’s constant = h = ?
When light of frequency 2 x 1015 Hz is incident on a metal surface, photoelectric current can be stopped by a retarding potential of 6 V. For light of frequency 1015 Hz the reverse potential is 2 V. Find Planck’s constant, work function and threshold frequency.
Given: Initial frequency = ν1 = 2 x 1015 Hz, initial stopping potential = Vs1 = 6 V, Final frequency = ν2 = 1015 Hz, Final stopping potential = Vs2 = 2 V, speed of light = c = 3 x 108 m/s, Charge on electron = e = 1.6 x 10-19 C.

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