Courses

# Electron Emission and Its Processes Class 12 Notes | EduRev

## Class 12 : Electron Emission and Its Processes Class 12 Notes | EduRev

The document Electron Emission and Its Processes Class 12 Notes | EduRev is a part of the Class 12 Course Physics Class 12.
All you need of Class 12 at this link: Class 12

Electron Emission Process: Fig: Electron emission process

When light is incident on a metal surface it was observed that electrons are ejected from a metal surface some times even when incredibly dim light such as that from stars and distance galaxies incident on it and some time electrons not comes out from the metal surface even high energetic or high intensity light falling on the metal surface.

This shows that the electron emission from a metal surface is not depends on the intensity of incident light but it is basically depends on the energy of the incident.

Photons no matters in number of photons are very less in a dim light, photo electric effect can be seen.

During the phenomenon of photoelectric effect one incident photon on metal surface can eject at most only one electron.

A photon is an energy packet which is fully absorbed not partially. Thus one photon can not be absorbed by more than one electron.

The minimum amount of energy of photon required to eject an electron out of a metal surface is called work function It is denoted by φ.

The work function depends on the nature of the metal.

1. The electron emission from a metal is only depends on the work function or energy of one photons.

2. But how many electrons comes out from the metal is depends on intensity of the falling light on energy of the light.

3. Energy of photon incident on metal will not necessarily cause emission of an electron even if its energy is more than work function. The electron after absorption may be involved in many other process like collision etc in which it can lose energy hence the ratio of no. of electrons emitted to the no. of photons incident on metal surface is less than unity.

1.6 Three Major Features of the Photoelectric effect cannot be explained in terms of the classical of the wave theory of light.

(a) The intensity problem: Wave theory requires that the oscillating electric field vector E of the light wave increases in amplitude as the intensity of the light beam is increased.
Since the force applied to the electron is eE, this suggests that the kinetic energy of the photoelectrons should also increased the light beam is made more intense. However observation shows that maximum kinetic energy is independent of the light intensity.

(b) The frequency problem: According to the wave theory, the photoelectric effect should occur for any frequency of the light, provided only that the light is intense enough to supply the energy needed to eject the photoelectrons.
However observations shows that there exists for each surface a characteristic cutoff frequency nth, for frequency less than nth, the photoelectric effect does not occur, no matter how intense is light beam.

(c) The time delay problem: If the energy acquired by a photoelectron is absorbed directly from the wave incident on the metal plate, the "effective target area" for an electron in the metal is limited and probably not much more than that of a circle of diameter roughly equal to that of an atom. In the classical theory, the light energy is uniformly distributed over the wavefront. Thus, if the light is feeble enough, there should be a measurable time lag, between the impinging of the light on the surface and the ejection of the photoelectron. During this interval the electron should be absorbing energy from the beam until it had accumulated enough to escape. However, no detectable time lag has ever been measured.

Now, quantum theory solves these problems in providing the correct interpretation of the photoelectric effect.

1.7 Threshold frequency and Threshold wavelength:

We have discussed that to start photoelectric emission, the energy of incident photon on metal surface must be more than the work function of the metal. If f is the work function of the metal then there must be a minimum frequency of the incident light photon which is just able to eject the electron from the metal surface. This minimum frequency or threshold frequency vth can be given as Threshold frequency nth is a characteristic property of a metal as it is the minimum frequency of the light radiation required to eject a free electron from the metal surface.

As the threshold frequency is defined, we can also define threshold wavelength for a metal surface. Threshold wavelength is also called cut off wavelength. For a given metal surface threshold wavelength is the longest wavelength at which photo electric effect is possible. Thus we have So for wavelength of incident light λ > λth, the energy of incident photons will become less then the work function of the metal and hence photoelectric effect will not start.

Thus for a given metal surface photoelectric emission will start at v > vth or λ > λth

1.7 Einstein relation:

Einstein suggested that the energy of photon (hv > φ) which is more than work function of a metal when incident on the metal surface is used by the electron after absorption in two parts.

(i) A part of energy of absorbed photon is used by the free electron in work done in coming out from the metal surface as work function.

(ii) The remaining part of the photon energy will be gained by the electron in the form of kinetic energy after ejection from the metal surface. If a light beam of frequency n (each photon energy = hn) is incident on a metal surface having work function f then for (hv > φ), we have ...(1)

In equation (1) the second terms on right hand side of equation is , which is the maximum kinetic energy of the ejected electron.

In practical cases whenever an electron absorbs a photon from incident light, it comes out from the metal surface if (hv > φ) but in process of ejection it may collide with the neighbouring electrons and before ejection it may loose some energy during collisions with the neighbouring electrons. In this case after ejection the kinetic energy of ejected electrons will be certainly less then (hv > φ). If we assume there are some electrons which do not loose any energy in the process of ejection, will come out from the metal surface with the maximum kinetic energy given as Thus all the ejected electrons from the metal surface may have different kinetic energies, distributed from 0 to .

Graph between Kmax and f: Fig: Graph between Kmax and f

Let us plot a graph between maximum kinetic energy Kmax of photoelectrons and frequency f of incident light. The equation between Kmax and f is,

Kmax = h f  – W

comparing it with y = mx + c, the graph between Kmax and f is a straight line with positive slope and negative intercept.

From the graph we can note the following points.

(i) Kmax = 0 at f = f0

(ii) Slope of the straight line is h, a universal constant. i.e., if graph is plotted for two different metals 1 and 2, slope of both the lines is same.

(iii) The negative intercept of the line is W, the work function, which is characteristic of a metal, i.e., intercepts for two different metals will be different. Further,

W2 > W1 Therefore, (f0)2 > (f0)1

Here f0 = threshold frequency as W = hf0

Ex. 2 The photoelectric threshold of the photo electric effect of a certain metal is 2750 Å. Find

(i) The work function of emission of an electron from this metal,

(ii) Maximum kinetic energy of these electrons,

(iii) The maximum velocity of the electrons ejected from the metal by light with a wavelength 1800 Å.

Ans: (i) Given that the threshold wavelength of a metal is λth =  2750 Å. Thus work function of metal can be given as φ = (ii) The energy of incident photon of wavelength 1800 Å on metal in eV is = 6.9 eV

Thus maximum kinetic energy of ejected electrons is

KEmax = E – φ = 6.9 - 4.52 eV = 2.38 eV

(iii) If the maximum speed of ejected electrons is vmax then we have = 2.38 eV

or = 9.15 X 105 m / s

Ex. 3 Light quanta with a energy 4.9 eV eject photoelectrons from metal with work function 4.5 eV. Find the maximum impulse transmitted to the surface of the metal when each electrons flies out.

Ans: According to Einstein's photoelectric equation

E = = hv - φ = 4.9 - 4.5 = 0.4 eV

If E be the energy of each ejected photo electron momentum of electrons is p = We know that change of momentum is impulse. Here the whole momentum of electron is gained when it is ejected out thus impulse on surface is Impulse = Substituting the values, we get

Maximum impulse = = 3.45 X10 -25 kg m / sec

Ex. 4 In a experiment tungsten cathode which has a threshold 2300 Å is irradiated by ultraviolet light of wavelength 1800 Å. Calculate

(i) Maximum energy of emitted photoelectron and

(ii) Work function for tungsten

(Mention both the results in electron-volts)

Given Plank's constant h = 6.6 X 10 -34 joule-sec, 1 eV = 1.6 X 10 -19joule and velocity of light c = 3 X 108 m/sec

Ans: The work function of tungsten cathode is = 5.4 eV

The energy in eV of incident photons is The maximum kinetic energy of ejected electrons can be given as

KEmax = E  - φ = 6.9 - 5.4 eV = 1.5 eV

Ex. 5 Light of wavelength 1800 Å ejects photoelectrons from a plate of a metal whose work functions is 2 eV. If a uniform magnetic field of tesla is applied parallel to plate, what would be the radius of the path followed by electrons ejected normally from the plate with maximum energy.

Ans: Energy of incident photons in eV is given as As work function of metal is 2 eV, the maximum kinetic energy of ejected electrons is KEmax = E  - φ = 6.9 – eV = 4.9 eV

If vmax be the speed of fasted electrons then we have = 4.9 X1.6 X10 -19 joule

or = 1.31 X 106 m/s

When an electron with this speed enters a uniform magnetic field normally it follows a circular path whose radius can be given by  or or r = 0.149 m

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

## Physics Class 12

204 videos|288 docs|125 tests

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;