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Mass Energy Equivalence

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The mass-energy equation is one of the critical underpinnings of Physics. German Physicist Albert Einstein set forth this popular regulation. This regulation expresses that mass and energy are comparative with one another. The mass-energy equation clarifies how energy can be changed over into mass and mass into energy. The hypothesis expresses that how much energy moved by an item is equivalent to its mass increased by the square of the speed of light.

As per Einstein’s Theory, comparable energy can be determined utilizing the mass (m) and the speed of light (c), which is stated as,

E = mc2

Mass-energy proportionality suggests that, despite the fact that the all-out mass of a framework changes, the all-out energy and force stay steady. Think about the impact of an electron and a proton. It annihilates the mass of the two particles yet produces a lot of energy as photons. The revelation of mass-energy proportionality demonstrated vital to the improvement of hypotheses of nuclear combination and splitting responses.

The Mass-Energy Formula

The relationship between the mass of the body and energy is represented by the below formula:

E = mc2


  • E is the Equivalent Kinetic Energy of the body,
  • m is the mass of the body or particle and
  • c is the speed of light (which is equal to 3 × 108 m/s).

The equation E=mc2 is supposed to characterize the energy of a molecule in its rest outline, which is meant by the capital letter E, as the result of mass, which is meant by the letter m, and the speed of light squared, which is c2. In the end, the mass of a molecule very still is equivalent to its energy, which is indicated by the letter E, partitioned by the speed of light squared, which is c2. Hence, the mass of a molecule very still is equivalent to its energy, which is meant by the letter E, separated by the speed of light squared, which is c2. We can say this because the speed of light is around 3 × 108 m/s, which is a big number in general terms.

We can argue that the mass and energy that were considered as separate entities were viewed as separate entities in the theories of physics previous to special relativity. In addition, we can state that the energy of a body at rest can be allocated an arbitrary number. Notwithstanding, in extraordinary relativity, it is expressed that the energy of a body very still is viewed as mc2. As a result, we can say that each body with a rest mass indicated by the letter m contains mc2 of “rest energy,” which is potentially available for conversion to other kinds of energy.

Derivation for Mass-Energy Formula

Here is the most general method to derive Einstein’s mass-energy equation as follows, Suppose an object is moving at a speed approximately equal to the speed of light. Now, a constant force is acting on it due to which in this case, the energy and momentum are come into play. Since, the force is uniform, then: 
The increase in Momentum of the object, p = mass (m) × velocity (v) of the object.

Also it is known that, the gained energy by this object, E = Force (F) × Displacement (c) through which the force acts.


E = F × c                                                                                                                                                                                            ………… (1)


The increased momentum = force × time through which force acts


Momentum = mass × velocity,

Hence, Force= m × c                                                                                                                                                                       ……………. (2)

Hence from the equation (1) and (2) we get,

E = mc2

Sample Problems 

Problem 1: One star in the universe is radiating with the energy of 7×1022 J/s. Determine the rate of mass decreasing of that star.


We know: E= mc2

ΔE = 7 × 1022 J/s

c = 3 × 108 m/s

From the above mass-energy equivalence formula we have:

ΔE = Δm × c2

7 × 1022 J/s = Δm × (3 × 108 m/s)2

Hence, decreasing rate of mass = 0.77 × 106 kg/s.

Problem 2: Suppose the velocity of a particle approaches the speed of light then what is the kinetic energy of that particle at that time?


From the mass-energy equivalence formula we have-

E = m / √(1- v2 /c2)                                                                                                                                                                              …………(1)

As v > c then v/c = 1

Putting value of v/c in equation (1) –

E = m / √(1-1) = m / 0

  = infinity

Hence kinetic energy will tends to infinity.

Problem 3: Deduce the rest energy of a proton (take the mass of the proton as 1.67 × 10-27 kg.


Since, m = 1.67 × 10-27 kg

c = 3 × 108 m/s

E = m × c2

Putting all values in the above equation we have :-

E = 1.67 × 10-27 kg × (3 × 108 m/s)2

   = 15.03 × 10-11 J

Problem 4: Determine the rest mass energy of 10 kg of water or any other substance.


The rest energy is defined as the amount of energy that is released when the complete mass of the given substance is converted entirely to energy. The rest mass of a substance is determined by using Einstein’s mass-energy equivalence formula.


E = m × c2

  = 10 kg × (3 × 108 m/s)2

  = 9 × 1017 J

Hence, the rest mass energy of water is 9 × 1017 J.

Problem 5: How much is the rest energy of an electron?


The rest energy of the electron can be calculated as,

E = mc2

where, m is the mass of the electron and is equal to 9.109 × 10-31 kg, and c is the speed of light and is equal to 3 × 108 m/s.

Therefore, substituting the values in the above equation:

E = 9.109 × 10-31 kg × (3 × 108 m/s)2

  = 8.198 × 10-14 J

  = 0.51 MeV.

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Last Updated : 25 Feb, 2022
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