
The energy of a photon is given by $E = hv$ where $v$ is the frequency of radiation. Use this equation to get the dimensional formula of $h$.
Answer
233.1k+ views
Hint: The expression provided in the question is known as Planck’s law and the constant is known as Planck’s constant. To find the dimension of the quantity h, the quantities energy E and frequency v must be resolved into the basic dimensions of the SI system through dimensional analysis.
Complete step by step answer:
The International System of Units has a set of seven basic quantities known as the fundamental quantities. They are:
A. Length
B. Mass
C. Second
D. Temperature
E. Current
F. Amount of substance
G. Luminous Intensity
The other physical quantities in nature are called derived quantities and they are derived from one or more of the above seven fundamental quantities.
Hence, any physical quantity can be expressed as a combination of these 7 basic quantities. This process is called dimensional analysis.
The Planck-Einstein relation is an equation based on Einstein's theory of photoelectric effect, that the energy of light exists in the form of discrete packets known as photons. Each photon has its energy proportional to the frequency of the light source. The energy of the photon is given by –
$E = hv$
where h = Planck’s constant whose value is equal to $6 \cdot 625 \times {10^{ - 34}}J{s^{ - 1}}$ and v = frequency.
From the above expression, we have –
$h = \dfrac{E}{v}$
The unit of energy is joule. 1 joule is equal to product of 1 newton and 1 metre.
$1J = 1Nm = 1\dfrac{{kgm}}{{{s^2}}}m = 1kg{m^2}{s^{ - 2}}$
Thus, we have expressed the unit joule in the basic fundamental units. Hence, the dimension of the energy is –
$E = [M L^2 T^{-2}]$
The unit of frequency is Hertz. 1 hertz is equal to the inverse of a second. Hence, the dimension of the frequency is –
$v = \dfrac{1}{{\sec }} = {\sec ^{ - 1}}$
$ \Rightarrow v = {\left[ T \right]^{-1}}$
Substituting these dimensions in the expression for h, we have –
$h = \dfrac{E}{v}$
$ \Rightarrow h = \dfrac{[M L^2 T^{-2}]}{[T^{-1}]}$
$ \Rightarrow h = [M L^2 T^{-1}]$
Hence, the dimension of the Planck’s constant, $h$ = $[M^1 L^2 T^{-1}]$.
Note: The concept of Planck’s relation is based on the famous photoelectric effect postulated by Einstein. Until this point of time, the light was considered to be a continuous wave. It was only Planck's law that proved that the energy transfer in electromagnetic radiation occurs, through discrete packets of energy and not continuous transfer of energy.
Complete step by step answer:
The International System of Units has a set of seven basic quantities known as the fundamental quantities. They are:
A. Length
B. Mass
C. Second
D. Temperature
E. Current
F. Amount of substance
G. Luminous Intensity
The other physical quantities in nature are called derived quantities and they are derived from one or more of the above seven fundamental quantities.
Hence, any physical quantity can be expressed as a combination of these 7 basic quantities. This process is called dimensional analysis.
The Planck-Einstein relation is an equation based on Einstein's theory of photoelectric effect, that the energy of light exists in the form of discrete packets known as photons. Each photon has its energy proportional to the frequency of the light source. The energy of the photon is given by –
$E = hv$
where h = Planck’s constant whose value is equal to $6 \cdot 625 \times {10^{ - 34}}J{s^{ - 1}}$ and v = frequency.
From the above expression, we have –
$h = \dfrac{E}{v}$
The unit of energy is joule. 1 joule is equal to product of 1 newton and 1 metre.
$1J = 1Nm = 1\dfrac{{kgm}}{{{s^2}}}m = 1kg{m^2}{s^{ - 2}}$
Thus, we have expressed the unit joule in the basic fundamental units. Hence, the dimension of the energy is –
$E = [M L^2 T^{-2}]$
The unit of frequency is Hertz. 1 hertz is equal to the inverse of a second. Hence, the dimension of the frequency is –
$v = \dfrac{1}{{\sec }} = {\sec ^{ - 1}}$
$ \Rightarrow v = {\left[ T \right]^{-1}}$
Substituting these dimensions in the expression for h, we have –
$h = \dfrac{E}{v}$
$ \Rightarrow h = \dfrac{[M L^2 T^{-2}]}{[T^{-1}]}$
$ \Rightarrow h = [M L^2 T^{-1}]$
Hence, the dimension of the Planck’s constant, $h$ = $[M^1 L^2 T^{-1}]$.
Note: The concept of Planck’s relation is based on the famous photoelectric effect postulated by Einstein. Until this point of time, the light was considered to be a continuous wave. It was only Planck's law that proved that the energy transfer in electromagnetic radiation occurs, through discrete packets of energy and not continuous transfer of energy.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

