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Planck’s Constant denoted by letter, ‘h’ is used to express the quantum nature of energy. The Planck’s constant gives the relationship between the energy of a photon and its energy. The nature of the particles and the waves can be understood from this constant on an atomic scale. Planck’s constant in SI units has a value of 6.62607015 x 10-34 J/Hz or Js.
Planck’s equation describes the relationship between temperature, spectral emissivity, and energy. It represents the spectral density of electromagnetic waves that are emitted by a black body at equilibrium at a given temperature when there is no net flow of energy or matter between the body and its environment. Planck’s Equation helps us calculate the energy of photons when the frequency is known.
| Table of Content |
Key Terms: Planck’s Law, Planck’s Constant, Planck’s Equation, Blackbody Radiation, Photons, Frequency, Electromagnetic Radiation
What is Planck's Law?
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Planck’s Law was given by Max Planck who discovered the energy transfer to be occurring in chunks, or quanta. A constant ‘h’ was assigned to evaluate the energy of the photons of a known frequency. The law states that:
The electromagnetic radiation from heated bodies that is not emitted as a continuous flow but is made up of quanta of energy, the size which involves a fundamental physical constant.
With an increase in temperature, the entire radiated energy increases, and due to this emitted spectrum shifts to a short wavelength. The mathematical equation for Planck’s Law is
\(B_{\lambda} (\lambda , T) = \frac{2hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k_B T}}- 1} \)
The spectral radiance of a body, Bλ, mentions the amount of energy it gives off as radiation of different frequencies. Planck’s equation describes the spectral radiance quantity radiated by a black body in equilibrium at a particular wavelength.
Planck’s Equation is given as –
\(E = hv = \frac{hc}{\lambda}\)
Where,
E = Energy
h = Plank constant
v = frequency
c = speed of light
λ = wavelength
- When the wavelength is known, the wave equation can be used to calculate the energy by using the frequency.
- He postulated that the energy of light is proportional to the frequency and there is a constant that relates these two factors called Planck's constant.
Read More:
| Important Topics Related to Planck’s Constant | ||
|---|---|---|
| Effects of Radiation | Electron Emission | Experimental Study of Photoelectric Effect |
| Compton Wavelength | Hadron | Quantum Theory of Light |
What is Planck's constant?
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Planck's constants is the relationship between the energy per quantum number of electromagnetic radiation to its frequency. The Planck's constant is multiplied by a photon frequency to have adequate photons. It is a physical constant and has importance in quantum physics. It is represented by the symbol ‘h’.
- The value of Planck's Constant (h) is equal to 6.63 x 10-34 J.s.
- As per the international system of units, the unit to describe in frequency is s-1.
- The Dimension of Planck’s constant is ML2T-1.
Value of Planck’s Constant in Different Units
The value of Planck’s constant h is determined experimentally. The table below summarises the value in different units.
| System of Units | Value |
|---|---|
| SI Units | 6.6260715 x 10-34Js |
| MKS Units (Meter-Second-Kilogram) | 4.135667662 x 10-15 eVs |
| Ep.tp | 2 pi |
Derivation of Planck's Law
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Every physical body in the universe is continuously emitting electromagnetic radiation and spectral radiation. B describes the emissive power per unit area and solid angle for specific radiation frequencies. The total radiation of body energy increases and as a result the peak of the emitted spectrum shifts to a shorter wavelength.
According to this spectral radiation of any physical body at a frequency and absolute temperature T is given by,
\(B(v, T) = \frac{2hv^3}{c^2} \frac{1}{e^{\frac{hv}{k_BT}}-1}\)
Where,
kB → Boltzmann Constant
h → Planck’s Constant
c → Speed of Light
The above equation can be simplified as follows.
\(B(v , T) = 2 v^3 \frac{1}{e^{\frac{v}{T}}-1}\)
Integrating the Spectral radiance per unit wavelength.
\(\int_{\lambda_1}^{\lambda_2} B(\lambda, T)d\lambda = \int_{v(\lambda_2)}^{v(\lambda_1)} B(v,T) dv =\int_{\lambda_1}^{\lambda_2} B(\lambda, T)\frac{dv}{d \lambda}d\lambda = \int_{\lambda_1}^{\lambda_2} -B(\lambda, T) \frac{dv}{d \lambda}d\lambda \)
On solving, we get,
\(B(\lambda, T) = -B(\lambda, T) \frac{dv}{d \lambda}\)
Using C = λv we can have the following formulae
\(B(\lambda, T) = \frac{2hc^2}{\lambda^2} \frac{1}{e^{\frac{hv}{k_BT}}-1}\)
Blackbody Radiations
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A blackbody is defined as a system that absorbs and emits all types of frequency. Planck's law closely describes the relationship of emitted radiation. It is also dependent on temperature therefore it is claimed to have thermal radiation. If there would be more a temperature in the body then there would be a high amount of radiation provided it emits wavelength.
Blackbody Radiation
- Planck's radiation is the highest radiation that a body can emit at equilibrium.
- A body emits thermal radiation that is highly infrared and invisible.
- If there is an increase in temperature then infrared radiation increases and can be felt as heat, and more visible radiation is emitted so the body glows visibly red.
- At higher temperatures, the body is slightly yellow or blue-white and emits actual amounts of short wavelength radiation.
- The surface of the sun (~6000 K) emits large amounts of both infrared and ultraviolet radiation, its emission peaks are seen in the visible spectrum.
Applications of Planck’s Constant
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Listed below are some of the important applications of Planck’s constant.
| Application | Equation |
|---|---|
| Calculating the spectral radiance of a body (blackbody radiation) | \(B(v, T) = \frac{2hv^3}{c^2} \frac{1}{e^{\frac{hv}{k_BT}}-1}\) |
| Photoelectric effect- Planck-Einstien Relation | E = hf |
| Atomic Structure- Energy of the nth orbit | \(E_n= \frac{-hc_0R_\infty}{n^2}\) |
| Uncertainity Priniciple- Position and Momentum Relation | \(\Delta x\Delta p\geq\frac{\frac{h}{2\pi}}{2}\) |
| Matter Wave Equation- de Brogllie Wavelength | \(\lambda= \frac{h}{p}\) |
Solved ExamplesQues: Wavelength of greenlight is 525nm. Calculate the frequency and energy of the green light? Soln: we know that C = λv Therefore, \(v=\frac{3\times10^8}{525}\) ∴v= 5.71 x 1014/s To calculate energy, we use the equation, Energy E= hv, where Planck’s constant h= 6.626 x 10-34 Js = 6.626 x 10-34 x 5.71 x 1014 = 3.78 x 10-19 J/Photon Ques: A red light has an energy of 2.84 x 10-19 J. Calculate the frequency. Soln: We know energy, E = hv Therefore, v= E/h = 2.84 x 10-19/ 6.626 x 10-34 = 4.29 x 1014 Hz or 4.29 x 1014 s-1 |
Notes on Dual Nature of Matter
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Provided below are some important concept notes on dual nature of matter.
Things to Remember
- The significance of Planck’s law is that it helps calculate the energy of the photon when the frequency is known.
- If the wavelength is known, then the frequency can be calculated and energy can be found by applying Planck’s equation.
- If Planck's constant is zero, then both momentum and position can be found at the same time.
- Planck radiation has a high intensity i.e at room temperature (~300 K), a body emits thermal radiation that is highly infrared and invisible.
- The value of Planck's Constant (h) is equal to 6.63 x 10³ J.s.
- As per the international system of units, the unit to describe in frequency is s-1.
- The Dimensional formula for Planck’s constant is ML²T¹.
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Previous Year Questions
- Mass of an electron when it moves with this speed is….[NEET 1990]
- An electron of mass m and charge e is accelerated from rest through a….[NEET 1996]
- An electron is accelerated from rest through a potential difference….[NEET 2020]
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- Two fast moving particles X and Y are associated with de Broglie wavelengths… [KEAM 2013]
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- When monochromatic radiation of intensity II falls on a metal surface…. [NEET 2010]
- An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is….[NEET 2019]
- If the Planck constant = 6.6 x 10-34 Js, the de Broglie wavelength… [BITSAT 2018]
- An electron is accelerated under a potential difference of 182 V… [BITSAT 2011]
- An electron of mass m with initial velocity V… [NEET 2018]
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- The kinetic energy of an electron get tripled then the de-Broglie… [VITEEE 2012]
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- An example for the best source of monochromatic light is... [JKCET 2019]
- An electron of mass m is accelerated by a potential difference V and the… [WBJEE 2016]
Sample Questions
Ques. What is Planck's law & Planck’s constant (2 marks)
Ans. Planck's law is stated as the energy of electromagnetic radiation is restricted to quanta, which further can't be divided and energy equal to the product of Planck’s constant and frequency of radiation. Planck’s constant is used for describing the particle behaviour and waves of atomic scale. Planck's constant is one of the main reasons for the development of quantum physics.
Ques. Red light has a wavelength of 630 nm. Find out the energy for the red light in joules. (3 marks)
Ans. To find out the frequency
We know that c = wavelength × frequency
Frequency = 3 × 108 / 630 × 10-9
Hence frequency = 4.7×1014
To find out the energy
h = 6.626 × 10-34
v = 4.7×1014
E = h× v
(6.626× 10-34) × (4.7×1014)
2.9× 10-19 j /photon
Ques. Why is LED used to determine Planck’s constant? (2 marks)
Ans. The LED is used for determining Planck’s constant because the LED has a different threshold voltage which is producing different photons and electrons. This voltage plus emission of electrons & photons helps in determining the plank constant.
Ques. How did Planck calculate the constant? (2 marks)
Ans. Planck found out this by calculating heat radiation given off by vibrating atoms, for this he had to predict that atom can only vibrate at some specific frequencies, which will be resulting in whole-number multiples of some base frequency which he termed as h, which is now known as Planck’s constant.
Ques. Greenlight has a wavelength of 550 nm. Find out the energy for the green light in joules. (2 marks)
Ans. We know that c= wavelength × frequency
Frequency = 3×108 / 550×10-9
Hence frequency = 5.4 × 1014
To find out the energy
h= 6.626 × 10-34
v = 5.4 × 1014
E= h × v
(6.626× 10-34) × (5.4×1014)
3.5× 10-19 j /photon
Ques. What is the emissive & absorptive power? (2 marks)
Ans. Emissive power is the amount of energy radiated per unit area of a body per unit time. It is denoted by (eλ). Absorptive power is defined as when the radiation of an incident ray is over the surface, it gets reflected, refracted & some part of it gets absorbed. It is denoted by (aλ).
Ques. What is Wein displacement law (2 marks)
Ans. Wilhelm Wien was the person who discovered this law. It is defined as when there is an increase in the temperature maximum value of the radiant energy emitted by the black body, it moves towards shorter wavelengths. The formula for Wien’s displacement law is as follows
λm x T= b
Ques. State two properties of photons. For a monochromatic radiation incident on a photosensitive surface, why do all photoelectrons not come out with the same energy? Give a reason for your answer. (CBSE 2017) (2 marks)
Ans. Two properties of photons :
- Photons are electrically neutral.
- Photon has an energy equal to hv
For a monochromatic radiation incident on a photosensitive surface, all photoelectrons do not come out with the same energy, because in addition to the work done to free electrons from the surface, different (emitted) photoelectrons need different amounts of work to be done on them to reach the surface.
Ques. Draw a plot showing the variation of photoelectric current with collector plate potential for two different frequencies, v1 > v2, of incident radiation having the same intensity. In which case will the stopping potential be higher? Justify your answer. (CBSE 2011) (2 marks)
Ans. Stopping potential is directly proportional to the frequency of incident radiation. The stopping potential is more negative for higher frequencies of incident radiation. Therefore, stopping potential is higher in v1.
Ques. Write Einstein’s photoelectric equation. State clearly how this equation is obtained using the photon picture of electromagnetic radiation. (CBSE 2011) (3 marks)
Ans.
\(hv = \phi_0 + K_{max}\)
This is Einstein’s photoelectric equation. Photoelectric emission is the result of the interaction of two particles—one a photon of incident radiation and the other an electron of photosensitive metal. The free electrons are bound within the metal due to restraining forces on the surface. The minimum energy required to liberate an electron from the metal surface is called work function of the metal. Each photon interacts with one electron. The energy hv of the incident photon is used up in two parts:
- A part of the energy of the photon is used in liberating the electron from the metal surface, which is equal to the work function ?0 of the metal and
- The remaining energy of the photon is used in imparting the K.E. of the ejected electron. By the conservation of energy Energy of the inefficient photon = maximum K.E. of photoelectron + Work function
\(hv = \frac{1}{2} mv^2_{max} + \phi _0\)
\(\implies K_{max} = \frac{1}{2} mv^2_{max} = hv - \phi_0 \)
\(= hv - hv_0 = h(v - v_0)\)
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