Question

What explanation was given by Einstein to explain photoelectric effect ? What do you understand by threshold frequency ?

Answer 1

Einstein’s Photoelectric Equation and Explanation of Experimental Results of Photoelectric Effect on the Basis of this Equation

In $$1905$$, Albert Einstein $$(1879-1955)$$ proposed a radically new picture of electromagnetic radiation to explain photoelectric effect. In this picture, photoelectric emission does not take place by continuous absorption of energy from radiation. Radiation of energy is built up of discrete units : the so called quanta of energy of radiation. Each quanta of radiant energy has energy $$hv$$, where $$h$$ is Planck’s constant and $$v$$ the frequency of light. In photoelectric effect, an electron absorbs a quanta of energy $$(hv)$$ of radiation. If the energy of quanta absorbed exceeds the minimum energy needed for the electron to escape from the metal surface (work function $$ \phi_0 $$), the light energy of radiation is given to the metal surface. This energy is spent into two ways. One part of energy is required to just eject the electron from metal surface and rest part of energy to provide the kinetic energy to the photon. So by the law of conservation of energy,

The energy of light = Work function + Kinetic energy

$$ hv = \phi + K_{max} $$ ..........(1)

When $$ v_{max} = 0 \Rightarrow \, K_{max} = 0 $$

then $$ v = v_0 $$

$$ \therefore \, hv_0 = \phi + 0 $$

$$ \therefore\, hv_0 = \phi $$ ...........(2)

$$ \therefore \, hv = hv_0 + K_{max} $$ ...........(3)

Equation (3) is called Einstein's photoelectric equation, basically it is a statement of law of conservation of energy for absorption of a single photon by a metal of work function $$ \phi_0 $$.

If mass of emitted electrons is $$m$$ and maximum velocity is $$v_{max} $$ then

$$ hv = \dfrac{1}{2} mv^{2}_{max} + \phi_{0} $$ ................(4)

If stopping potential is $$ V_0 , K_{max} = eV_0$$

$$ hv = eV_0 + \phi_0 $$ ..............(5)

Equations (4) and (5) are the other forms of photoelectric equation. Now we see how this equation accounts in a simple and elegant manner for all the observations on photoelectric effect.

(i) According to equation (3), $$ K_{max} $$ depends linearly on frequency of incident radiation (v), and is independent of intensity of radiation, in the agreement with observation. This has happened because in Einstein’s picture, photoelectric effect arises from the absorption of a single quantum of radiation by a single electron. The intensity of radiation (that is proportional to the number of energy quanta per unit area per unit time) is irrelevant to this basic process.

(ii) Since $$ K_{max} $$ must be non-negative, equation (3) implies that photoelectric emission is possible only if

$$ hv > \phi_0$$

or $$ hv > hv_0 $$ , where

$$ v_0 = \dfrac{\phi_0}{h} $$ .........(6)

Equation (6) shows that the greater the work function $$ \phi_0$$ , the higher the minimum or threshold frequency $$ v_0$$ needed to emit photoelectrons. Thus there exists a threshold frequency below which no photoelectric emission is possible, no matter how intense the incident radiation is or how large it falls on the surface.

(iii) In this picture, intensity of radiation as noted above, is proportional to the number of energy of quanta per unit area per unit time. The greater the number of energy quanta available, the greater is the number of electrons absorbing the energy quanta and greater, therefore, is the number of electrons coming out from the metal (for $$ v > v_0$$). This explains why, for $$ v > v_0 $$ , photoelectric current is proportional to intensity.

(iv) In Einstein’s picture, the basic elementary process involved in photoelectric effect is the absorption of light quantum by an electron. This process is instantaneous. Thus whatever may be the intensity i.e. the number of quanta of radiation per unit area per unit time, photoelectric emission is instantaneous. Intensity only determines how many electrons are able to participate in the elementary process (absorption of a light quantum by a single electron), and therefore, the photoelectric current.

$$ \because \, K_{max} = eV_0 $$, so the photoelectric equation (3) can be written as

$$ eV_0 = hv - \phi_0 $$ ; for $$ v \geq v_0 $$

or $$ V_0 = \left ( \dfrac{h}{e} \right ) v - \dfrac{\phi_0}{e} $$ ......(7)

This is an important result. It predicts that the $$V_0$$ versus $$v$$ curve is a straight line with slope = $$ (h/e) $$, independent of the nature of the material. During $$1906 -1916$$, Millikan performed a series of experiments on photoelectric effect, aimed at disproving Einstein’s photoelectric equation. He measured the slope of the straight line obtained for sodium, similar to that shown in figure. Using the known value of $$e$$, he determined the value of Planck’s constant $$h$$. This value was close to the value of Planck’s constant $$ (= 6.626 \times 10^{-34}\, J-s) $$ determined in an entirely different reference. In this way, in $$1916$$, Millikan traced the validity of Einstein’s photoelectric equation instead of disproving it.

The successfully explanation of photoelectric effect using the hypothesis of light energy or quanta and the experimental determination of values of $$h$$ and in agreement with values obtained from other experiments, led to the acceptance of Einstein’s picture of photoelectric effect. Millikan verified photoelectric equation with great precision, for a number of alkali metals over a wide range of radiation frequencies.