## What is Planck's Constant 'h'?

Max Planck, a German scientist, was interested in understanding the electromagnetic radiation emitted by a blackbody. A body emits and absorbs electromagnetic radiation incident on it. But a blackbody can absorb and emit all the incident radiation on it. A blackbody emits the incident radiation at a constant temperature. Planck was keen on understanding the spectrum of this emission or the amount of energy radiated at a specific wavelength. Before Planck, it was assumed that the energy associated could take any value and the energy emission is continuous. But it was later observed that more energy is produced at higher frequencies; in other words, a heated body would radiate more blue than red. This was contrary to what was observed before. Planck later theorized that if he assumes energy to be radiated in packets, the model can produce the same outcome as in Black Body radiation without violating already proven physics principles.

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He then derived the value of a constant, later known as Planck's constant 'h,' which relates the energy of a photon to its frequency. When Planck's constant is multiplied by a photon's frequency, the result is the specific photon's energy. The value of Planck's constant 'h' is 6.626070 * 10^{-34} JS. Planck calculated this value of h from the experimental data obtained from black body radiations.

### Value of 'hc' and its Importance

Planck's constant' h' and speed of light' c' can be used to find a photon's energy. A photon's energy is directly proportional to the frequency and thus is inversely proportional to its wavelength. Higher photon frequency denotes higher energy, and the longer is the photon particle's wavelength, the lower is the energy.

The value of 'hc' is 1.986445 * 10^{-25} J m. The equation for the energy of a photon is given as E=hc/λ. Here E is the energy of the photon, h is Planck's constant, c is the speed of light, and λ denotes the wavelength of the photon. The value of 'hc' in eV comes out to be 1.23984193 * 10^{-6} eV.m. The value of 'hc' is also used to find out the energy of a specific atomic orbit by using the formula, E= - hcR/n^{2}

### Role of hc in Photoelectric Effect

hc plays a vital role in the photoelectric effect. But to understand this first we have to know what is photoelectric effect? Photoelectric Effect is when electrons are ejected out from a metal surface when light is incident on it. It was then believed that the more the intensity of incident light, the more energy it possesses regardless of the color of the light. The light was considered and treated to be a wave, but the wave phenomena didn't explain the photoelectric effects of light. The photoelectrons emitted during the phenomenon of the photoelectric effect have a definite and measurable kinetic energy. The Kinetic energy also depends linearly on the frequency along with the intensity. If the frequency is too low, there is no discharge of photoelectrons. A threshold frequency is needed to take the electrons in atoms of their orbit. After the threshold frequency is crossed, the number of photoelectrons emitted increases with the intensity of radiation.

### Constant h in Atomic Structure

To overcome Rutherford's classical model's drawbacks, Neils Bohr introduced the first quantized model of an atom in 1913. In classical mechanics, particles spiraling around should emit electromagnetic radiation, thus losing energy and gradually spiral down into the nucleus. To solve this paradox, Bohr took reference from Planck's work and postulated that an electron would spiral around the nucleus only in orbits having pre-defined energies. The formula gives the energy of a specific orbit, E= -hcR/n^{2}, where R is the experimentally determined Rydberg's constant and 'n' is the orbit number. When the electron reaches the lowest energy level, i.e., n=1, it cannot move any closer to the nucleus. The constant 'h' is also used in the uncertainty principle given by Heisenberg.

### Overall Significance of h

Planck's Constant establishes a relation between photon energy and frequency and gives a quantitative value to light in the microscopic context where classical mechanics fails. The order of Planck's Constant, in a way, says that macroscopic bodies around us are made of billions and trillions of small microscopic bodies, which are in turn governed by a new set of laws altogether. Planck's Constant had a significant effect on our understanding of reality and life itself. The value also has an essential role in the construction of transistors, Integrated Circuits, and chips. Planck's quantization of energy was a revolutionary step to the destination of quantum mechanics, i.e., the way particles behave at the minor level.

### Points to Remember

hc = 12,400 eV

Å = 1240 eV

nm = 1240 MeV

fm

Dimension of hc = [M

^{1}L^{3}T^{-2}]

### Solved Numerical Type Questions

Q1. The work function of a metal is 3 eV. Will this metal give photoelectric emission for incident radiation of wavelength 350 nm? (h=6.63 x 10^{-3} J s, c = 3 x 10^{8} m/sec)

Sol. The energy of the incident radiation

E=hc/λ

= (6.63 x 10^{-34} x 3 x 10^{8})/(350 x 10^{-9})

= 0.0568 x 10^{-17} J

= 5.68 x 10^{-19} / 1.6 x 10^{-19}

= 3.55 eV

Since the work function of the given metal is 3 eV. Therefore, when the photon of high energy incident on the metal, photoemission occurs.

2. A radio transmitter operates on a wavelength of 2500 m at a power of 500 kW. Find the energy of radio photons in joules. (h = 6.64 x 10^{-34} J s, c = 3 x 10^{8} m/sec)

Solution. The energy of a photon emitted by a radio transmitter that operates on a wavelength of 2500 m is given by

E = hc/λ

= (6.64 x 10^{-34} J s x 3 x 10^{8} m/s)/(2500 m)

= 0.7968 x 10^{-28} J

So, the energy of photoelectron emitted from a radio transmitter is 79.68 x 10^{-30} J.

**1. What is the use of h in physics for a student?**

h, i.e., Planck's constant is needed to calculate the energy of photoelectrons when the wavelength or frequency of it is provided in the question, using the formula E=hc/λ. When the frequency is given instead of wavelength, the frequency can be a substitute for c/λ. Hence the formula becomes E= hf. Planck's Constant is also used to calculate the energy of a certain orbit in an atom, and the formula is given by E= - hcR/n^{2}.

**2. What is the unit of Planck's constant?**

In the MKS system, the unit of Planck's constant is Joule-s. It can be understood from the formula, E=hf, where f is the frequency. h =E/f, the units of E being Joule and that of f is s^{-1}. Thus, we arrive at Js .

**3. What are the values of hc in M.K.S. unit systems?**

To evaluate the value of hc in M.K.S.and S.I. unit systems, we have to find the values of Plank’s constant h and speed of light c in both unit systems.

In M.K.S. system: h = 4.135667662 x 10^{-15} eV-s, c ≃ 3 x 10^{8} m/s

∴ hc = 1.23984193 × 10^{-6} eV-m

In S.I. system: h = 6.6260715 x 10^{-34} J.s, , c ≃ 3 x 10^{8} m/s

∴ hc = 1.98644568×10^{-25} J.m

**4. What are the applications of hc?**

hc is a physical constant that is widely used in the study of the photoelectric effect, especially in Planck’s constant. Also, it is used to find the energy of the n^{th} orbit in an atomic structure.

**5. Find the dimension of hc.**

Dimension of Planck’s constant h = [ML^{2}T^{-1}]

Dimension of Speed of light c = [M^{0}L^{1}T^{-1}]

∴ dimension of hc = [ML^{2}T^{-1}] x [M^{0}L^{1}T^{-1}] = [M^{1}L^{3}T^{-2}]