Print ViewDoes the Photoelectric Effect Depend on the Properties of the Incident Light?

Question

INTRO:

Light striking a metal surface causes electrons to be emitted from the metal via the photoelectric effect.

PART A:

In a particular experiment to study the photoelectric effect, the frequency of the incident light and the temperature of the metal are held constant. Assuming that the light incident on the metal surface causes electrons to be ejected from the metal, what happens if the intensity of the incident light is increased?

Check all that apply.

1) The work function of the metal decreases.

2) The number of electrons emitted from the metal per

second increases.

3)The maximum speed of the emitted electrons increases.

4) The stopping potential increases.

Answer 1

Light is emitted as photons which have energy:-

       Energy of 1 photon   =   Planck's constant (h)  x  Frequency (f)

                                         E     =     h*f

It requires energy to remove an electron from a metal surface.  This is called the "Work Function" (W).

The relevant equation is:-

    Kinetic Energy of emitted electron   =   Photon Energy   -   Work Function

            K.E. of electrons    =    h*f   -   W

We can see that electrons are emitted (  i.e. have +ve K.E.)  if  h*f  >  W

    The answers.

    ==========

    1)    Wrong. The Work function of a metal is constant for that metal. You have to change the metal to get a different Work Function.

    2)  Correct..  When the intensity of light increases, you get more photons. But  each photon has the same energy as before.  

     3)  Wrong.   If you look at the equation, to get faster electrons with  the same metal  (more K.E.), you have to increase the photon energy (h*f). Since "h" is constant, the only way is to use light of higher frequency ( i.e. shorter wavelength).

     4)  Wrong.  The stopping potential is used in an experiment to find what the K.E. of the emitted electrons is.

The stopping potential prtevents emitted electrons reaching the anode.

Answer 2

the correct answer is 2) in accordance with

hxfrequency=kinetic energy of the ejected electron+work function of the metal.

which holds for the interaction of a single photon with a single electron. increasing the intensity more photons interact with more electrons and so the number of electrons emitted from the metal per second increases.

Answer 3

The photons of the light beam have a characteristic energy determined by the frequency of the light. In the photoemission process, if an electron absorbs the energy of one photon and has more energy than the work function, it is ejected from the material. If the photon energy is too low, the electron is unable to escape the surface of the material.

Increasing the intensity of the light beam increases the number of photons in the light beam, and thus increases the number of electrons emitted without increasing the energy that each electron possesses. Thus the energy of the emitted electrons does not depend on the intensity of the incoming light, but only on the energy of the individual photons.

Experimental results of the photoelectric emission

For a given metal and frequency of incident radiation, the rate at which photoelectrons are ejected is directly proportional to the intensity of the incident light.

For a given metal, there exists a certain minimum frequency of incident radiation below which no photoelectrons can be emitted. This frequency is called the threshold frequency.

Above the threshold frequency, the maximum kinetic energy of the emitted photoelectron is independent of the intensity of the incident light but depends on the frequency of the incident light.

The time lag between the incidence of radiation and the emission of a photoelectron is very small, less than 10-9 second

Energy of photon = Energy needed to remove an electron + Kinetic energy of the emitted electron

Algebraically:

h_f = phi+E

phi = is work function  the minimum energy required to remove a delocalised electron from the surface of any given metal )

E = 1/2  m v^2  KE of emitted electrons

h is Planck's constant

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