Assumptions used:
1. The particles/oscillators near the surface of the blackbody which emits the blackbody radiation can only have discrete values of energy, En: En=nhf, where n is a positive interger, f is the frequency of the oscillating particle, h is the Planck’s constant.
- Particles can only have discrete values of energy. Each energy value corresponds to a quantum state of the particle, represented by the corresponding quantum number.
2. In order for a particle to transit from one quantum state to another, the particle has to absorb or emit energy. The difference in the energy between the initial and the final state in the transition must be absorbed or emitted as a SINGLE quantum of energy. Smallest amount of energy involved is when the transition occurs between two adjacent states: E = hf.
- Energy is quantized in steps of hf
- One quanta of energy or radiation is known as a photon
Planck’s Law Of Blackbody Radiation:
$$I \left( v, T \right) dv = \left( \frac{2hv^{3}}{c^{2}} \right) \frac{1}{e^{\frac{hv}{kt}} – 1} dv$$
, where
I(ν,T) dν is the amount of energy per unit surface area per unit time per unit solid angle emitted in the frequency range between ν and ν + dν by a black body at temperature T;
h is the Planck constant;
c is the speed of light in a vacuum;
k is the Boltzmann constant;
ν is frequency of electromagnetic radiation; and
T is the temperature in kelvins.
Deriving the momentum of a photon:
Using E = pc and E = hf,
$$\begin{eqnarray*} E &=& hf \\ E &=& h \left( \frac{c}{\lambda} \right), \, \text{since} \, f = \frac{c}{\lambda} \end{eqnarray*}$$
Substituting E = pc into $E = h \left(\frac{c}{λ} \right)$:
$$\begin{eqnarray*} h \left( \frac{v}{\lambda} \right) &=& pv \\ p &=& \left( \frac{h}{\lambda} \right) \end{eqnarray*}$$
