Planck's quantum theory

Quantization of Energy - Why Energy Comes in Bits?

When the energy is emitted and absorbed in discrete quantities, these discrete quantities are called Quanta, which is the quantization of energy. This term was later termed photons by Albert Einstein.

In short, radiant energy is emitted or absorbed not continuously but discontinuously in the form of small discrete packets of energy. Each such packet of energy is called a 'quantum'. In case of light, the quantum of energy is called a 'photon'. The energy of each quantum is directly proportional to the frequency ($\nu$) of radiation and is expressed by the following equation:

$E=h v=\frac{h c}{\lambda}$

where h is the Planck's constant, and it has a value equal to $6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}$

The total amount of energy emitted or absorbed by a body will be some whole number quanta. Hence,

$E=n h v$ Where, n is any integer.

Note: The energy possessed by one mole of quanta (or photons), i.e., Avogadro's number($
N_0
$) of quanta is called one einstein of energy, i.e.,

1 Einstein of energy $
E=N_0 h \nu=N_0 h \frac{c}{\lambda}
$

Particle Nature of Electromagnetic Radiation: Planck's Quantum Theory

Till now the Electromagnetic radiations were thought to show wave like character. And no doubt, wave character was even able to clearly justify some of the phenomenon like diffraction and interference.

Diffraction: It is bending of wave around an obstacle.

Interference: It is combination of two waves of the same or different frequencies to give a new wave whose distribution at each point in space is the algebraic or vector sum of disturbances at that point resulting from each interfering wave.

But the wave nature of light could not justify some other obsevations like:

• Black-body radiation
• Photoelecrtic effect
• Variation of heat capacity of solids as a function of temperature.
• Line spectra of atoms with special reference to hydrogen

Explanation of Black-body Radiation

When some soild substance is heated, the atoms of the substance are set into oscillation and emit radiation of frequency, $\nu$. Now, as heating is continued, more and more energy is being absorbed by the atoms and they emit radiations of higher and higher frequency. As red light has minimum frequency and yellow has higher frequency, therefore, the body on heating becomes first red, then yellow and so on.

Also Read:

• Photoelectric Effect
• Bohr's Model Of An Atom

Recommended Topic Video on ( Planck's Quantum Theory)

Let's try to solve the question based on the above formula,

Example 1:

Which one of the following is not characteristic of Planck's quantum theory of radiation:

1) The energy is not absorbed or emitted in whole numbers or multiple quantums.

2)Radiations are associated with energy.

3) Radiation energy is not emitted or absorbed continuously but in the form of small packets called quanta.

4) This magnitude of energy associated with a quantum is proportional to the frequency.

Solution:

According to Planck's Quantum Theory,

Energy is absorbed or emitted not in a continuous manner, but in the form of small packets called Quanta.

The energy associated with the quanta is proportional to the frequency.

Hence, the answer is the option (1).

Example 2: Which of the following statements is false?

1) Photon has momentum as well as wavelength.

2) The splitting of spectral lines in an electrical field is called the Stark effect.

3) The Rydberg constant has a unit of energy

4) The frequency of emitted radiation from a black body goes from a lower wavelength to a higher wavelength as the temperature increases.

Solution

As we learned in

The energy (E) of a quantum of radiation -

$
E=h \nu
$

Where h is plank's constant and $\nu$ is the frequency.

When the temperature is increased, the black body emits high-energy radiation, from a higher wavelength to a lower wavelength.

Hence, the answer is an option (4).

Example 3:The energy of one quantum of light with a wavelength of $6500 \mathrm{~A}^{\circ}\left(1 \mathrm{~A}^{\circ}=10^{-10} \mathrm{~m}\right)$ is

1)$5 \times 10^{-18} \mathrm{~J}$

2) $3.06 \times 10^{-19} \mathrm{~J}$

3) $8.02 \times 10^{-20} \mathrm{~J}$

4) $9.0 \times 10^{-24} \mathrm{~J}$

Solution

As we learn

Planck’s Quantum Theory -

Atoms and molecules could emit (or absorb) energy only in discrete quantities known as quanta, and not continuously.

The energy of one quantum of light

$\begin{aligned} & \mathrm{E}=\mathrm{hc} / \lambda=\left(6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right) \times\left(3 \times 10^{-8} \mathrm{~m} / \mathrm{s}\right) /\left(6.5 \times 10^{-7}\right) \\ & =3.06 \times 10^{-19} \mathrm{~J}\end{aligned}$

Hence, the answer is the option (2).

Example 4: According to Planck's quantum theory, the energy of a quantum of radiation is given by:

1) $
E=h v
$
2) $E=h c \lambda$
3) $E=h c / 2 \lambda$
4) $E=h \lambda$

According to Planck's quantum theory, the energy of a quantum of radiation is directly proportional to its frequency. This is given by the equation $
E=h v
$, where E is the energy of the quantum, h is Planck's constant $\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)$, and v is the frequency of the radiation.

Hence, the answer is the option (1).

Example 5: What is the wavelength of a photon with energy $3.0 \times 10^{-19} \mathrm{~J}$, according to Planck's quantum theory?
1) $1.15 \times 10^{-7} \mathrm{~m}$
2) $2.50 \times 10^{-7} \mathrm{~m}$
3) $5.56 \times 10^{-7} \mathrm{~m}$
4) $9.23 \times 10^{-7} \mathrm{~m}$

Solution

According to Planck's quantum theory, the energy of a photon is given by the equation $\mathrm{E}=\mathrm{hc} / \lambda$, Where E is the energy of the photon, h is Planck's con$\left(6.626 \times 10^{-34} \mathrm{~J}-{~s}\right.$),c is thed the speed of light $\left(3.0 \times 10^8 \mathrm{~m} / \mathrm{s}\right)$, λ is the wavelength of the photon.

Rearranging this equation gives

$\lambda=\mathrm{hc} / \mathrm{E}$

Plugging in the given values, we get$\begin{aligned} & \lambda=\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\left(3.0 \times 10^8 \mathrm{~m} / \mathrm{s}\right) /\left(3.0 \times 10^{-19}\right) \\ & J \approx 2.50 \times 10^{-7} \mathrm{~m}\end{aligned}$

Hence, the answer is the option (2).

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Conclusion

We can conclude that Planck's quantum theory gave a whole new perspective and also a whole new paradigm shift in science, and we understood energy absorption and emission. The black body radiation which was initially not cleared by classical physics is much clearer in Quantum mechanics. Planck's postulate or Planck's Hypothesis explains the energy emitted and absorbed are in discrete quantities and termed as Quantization of energy and the term quanta termed Photons by Albert Einstein. The nature of emission or absorption of radiation from hot bodies (black-body radiation). The ejection of electrons from a metal surface when radiation strikes it called the photoelectric effect. Line spectra of atoms with special reference to hydrogen(Bohr's model of hydrogen).