Photoelectric Effect

When a beam of light of sufficiently high frequency is allowed to strike a metal surface, electrons are ejected from the metal surface. This phenomenon is known as photoelectric effect and the ejected electrons are called photoelectrons.
Example- When UV light shines on group IA elements as in the apparatus shown in fig-1, the photoelectric effect occurs.

With the help of this photoelectric apparatus, the following observations can be made-
1. In increase in intesity of incident light does not increase the energy of photoelectrons. It mearly increases their rate of emission.
2. The kinetic energy of photoelectrons increases linearly with the frequency of the incident light. If the frequency is decreased below a certain critical value (i.e. Threshold frequency, ν^o) no electrons are ejected at all. Fig-2.

"The classical physics predicts that the kinetic energy of photoelectrons should depends upon the intensity of light and not on the frequency".

Explanation of Photoelectric Effect

Einstein Explanation

1. A photon of incident light transmits its energy (hν) to an electron in the metal surface which escapes with kinetic energy 1/2mv². The greater intensity of incident light merely implies greater number of photon each of which releases one electron. this increases the rate of emission of electrons while the kinetic energy of individual photons remains unaffected. Fig-3

2. The energy of photon (hν) is Proportional to the frequency of the incident light. The frequency which provides enough energy just to release the electron from the metal surface, will be the threshold frequency ν^o. For frequency less than ν^o, no electrons Will be emitted.
For higher frequeny, ν > ν^o, a part of the energy goes to loosen the electron and remaining for imparting kinetic energy to the Photoelectron. Thus,
hν = hν^o + 1/2 mv²
where hν is the energy of the incoming photons, hν^o is the minimum energy for an electron to escape from the metal and 1/2mv² is the kinetic energy of the photoelectron.
hν^o is constant for a particular solid and is designated as (w), the work function.
1/2 mv² = hν − w
This is the equation of straight line that was experimentally obtained in fig.-2.

What is the minimum energy that photons must possess in order to produce photoelectric effect with Pt-metal. The threshold frequency for Pt is 1.3 x 10²⁵ seconds⁻¹.

The ν^o is the lowest frequency that photons may possess to produce the photoelectric effect. The energy corresponding to this frequency is the minimum energy.
E = hν^o
or, E = (6.626 x 10⁻²⁷ erg sec) x (1.3 x 10²⁵ sec⁻¹)
or, E = 8.6 x 10⁻¹² erg.

A photocell is receiving light from a source placed at a distance of 1m. If the same source is to be placed at a distance of 2 m, then the ejected electron
a. Moves in one-fourth energy as that of the initial energy
b. Moves with one-fourth momentum as that of the initial momentum
c. Will be half in number
d. Will be one-fourth in number

The minimum energy required to remove an electron is called
a. Stopping potential
b. Kinetic energy
c. Work function
d. None of these

How does the intensity affect the photoelectric current?
a. As intensity increases, the photoelectric effect increases
b. As the intensity increases, the photoelectric effect decreases
c. As the intensity decreases, the photoelectric effect becomes twice
d. No effect

What is the effect of intensity on the stopping potential?
a. As intensity increases, stopping potential increases linearly
b. As intensity increases, stopping potential decreases linearly
c. As intensity decreases, stopping potential increases exponentially
d. No effect

The photoelectric emission could be explained by the
a. Wave nature of light
b. Particle nature of light
c. Dual nature of light
d. Quantum nature