Advertisements
Advertisements
Question
Light of frequency 7.21 × 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Advertisements
Solution
Frequency of the incident photon, v = 7.21 × 1014 Hz
Maximum speed of the electrons, ν = 6.0 × 105 m/s
Planck’s constant, h = 6.626 × 10−34 Js
Mass of an electron, m = 9.1 × 10−31 kg
For threshold frequency v0, the relation for kinetic energy is written as:
`1/2 mν^2 = h(v - v_0)`
`v_0 = v - (mν^2)/(2h)`
= `7.21 xx 10^14 - ((9.1 xx 10^(-31) ) xx (6 xx 10^(5))^2)/(2 xx (6.626 xx 10^(-34)))`
= 7.21 × 1014 − 2.472 × 1014
= 4.738 × 1014 Hz
Therefore, the threshold frequency for the photoemission of electrons is 4.738 × 1014 Hz.
RELATED QUESTIONS
The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
(a) A monoenergetic electron beam with electron speed of 5.20 × 106 m s−1 is subject to a magnetic field of 1.30 × 10−4 T normal to the beam velocity. What is the a radius of the circle traced by the beam, given e/m for electron equals 1.76 × 1011 C kg−1?
(b) Is the formula you employ in (a) valid for calculating the radius of the path of a 20 MeV electron beam? If not, in what way is it modified?
Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?
What is so special about the combination e/m? Why do we not simply talk of e and m separately?
If light of wavelength 412.5 nm is incident on each of the metals given below, which ones will show photoelectric emission and why?
| Metal | Work Function (eV) |
| Na | 1.92 |
| K | 2.15 |
| Ca | 3.20 |
| Mo | 4.17 |
Two neutral particles are kept 1 m apart. Suppose by some mechanism some charge is transferred from one particle to the other and the electric potential energy lost is completely converted into a photon. Calculate the longest and the next smaller wavelength of the photon possible.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
A light beam of wavelength 400 nm is incident on a metal plate of work function 2.2 eV. (a) A particular electron absorbs a photon and makes two collisions before coming out of the metal. Assuming that 10% of the extra energy is lost to the metal in each collision, find the kinetic energy of this electron as it comes out of the metal. (b) Under the same assumptions, find the maximum number of collisions the electron can suffer before it becomes unable to come out of the metal.
A horizontal cesium plate (φ = 1.9 eV) is moved vertically downward at a constant speed v in a room full of radiation of wavelength 250 nm and above. What should be the minimum value of v so that the vertically-upward component of velocity is non-positive for each photoelectron?
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
Plot a graph to show the variation of stopping potential with frequency of incident radiation in relation to photoelectric effect.
Work function of aluminium is 4.2 eV. If two photons each of energy 2.5 eV are incident on its surface, will the emission of electrons take place? Justify your answer.
The stopping potential in an experiment on photoelectric effect is 1.5V. What is the maximum kinetic energy of the photoelectrons emitted? Calculate in Joules.
In the case of a photo electric effect experiment, explain the following facts, giving reasons.
The wave theory of light could not explain the existence of the threshold frequency.
The electromagnetic theory of light failed to explain ______.
In photoelectric effect, the photoelectric current
An increase in the intensity of the radiation causing photo-electric emission from a surface does not affect the maximum K.E. of the photoelectrons. Explain.
