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Dual Nature of Radiation and Matter — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsDual Nature of Radiation and Matter5 Marks
Question
Derive Einstein’s photoelectric equation $\frac{1}{2}\text{m}\text{v}^2=\text{h}\nu-\text{h}\nu_0.$
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Answer

Einstein’s Explanation of Photoelectric Effect:

Einstein’s Photoelectric Equation:

Einstein explained photoelectric effect on the basis of quantum theory. The main points are,

  1. Light is propagated in the form of bundles of energy. Each bundle of energy is called a quantum or photon and has energy hν where h = Planck’s constant and ν = frequency of light.
  2. The photoelectric effect is due to collision of a photon of incident light and a bound electron of the metallic cathode.
  3. When a photon of incident light falls on the metallic surface, it is completely absorbed. Before being absorbed it penetrates through a distance of nearly 10-8m (or $100\mathring{\text{A}}$). The absorbed photon transfers its whole energy to a single electron. The energy of photon goes in two parts: a part of energy is used in releasing the electron from the metal surface (i.e., in overcoming work function) and the remaining part appears in the form of kinetic energy of the same electron.
If ν be the frequency of incident light, the energy of photon = hν. If W be the work function of metal and EK the maximum kinetic energy of photoelectron, then according to Einstein’s explanation.

$\text{h}\nu=\text{W}+\text{E}_\text{K}$ or $\text{E}_\text{K}=\text{h}\nu-\text{W}\dots(\text{i})$

This is called Einstein’s photoelectric equation. If $\nu_0$ be the threshold frequency, then if frequency of incident light is less then $\nu_0$ no electron will be emitted and if the frequency of incident light be $\nu_0$ then EK = 0; so from equation (i)

$0=\text{h}\nu_0-\text{W}$ or $\text{W}=\text{h}\nu_0$

If $\lambda_0$ be the threshold wavelength, then $\nu_0=\frac{\text{c}}{\lambda_0},$

where c is the speed of light in vacuum

$\therefore$ Work function $\text{W}=\text{h}\nu_0=\frac{\text{hc}}{\lambda_0}\dots(\text{ii})$

Substituting this value in equation (i), we get

$\text{E}_\text{K}=\text{h}\nu-\text{h}\nu_0$

$\Rightarrow\ \frac{1}{2}\text{mv}^2=\text{h}\nu-\text{h}\nu_0\dots(\text{iii})$

This is another form of Einstein's photoelectric equation.

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