$M_p$ denotes the mass of a proton and $M_n$ denotes the mass of a neutron. A given nucleus of binding energy $B$ contains $Z$ protons and $N$ neutrons. The mass $M(N,Z)$ of the nucleus is given by (where $c$ is the speed of light) (a) $M(N,Z)=NM_n+ZM_p-Bc^2$ (b) $M(N,Z)=NM_n+ZM_p+Bc^2$ (c) $M(N,Z)=NM_n+ZM_p-\dfrac{B}{c^2}$ (d) $M(N,Z)=NM_n+ZM_p+\dfrac{B}{c^2}$

Mp denotes the mass of a proton and Mₙ denotes the mass of a neutron. A given nucleus of binding energy B cont

I Multiple Choice Questions · Q12

M_p

(a)

M(N,Z)=NM_n+ZM_p-Bc^2

(b)

M(N,Z)=NM_n+ZM_p+Bc^2

(c)

M(N,Z)=NM_n+ZM_p-\dfrac{B}{c^2}

(d)

M(N,Z)=NM_n+ZM_p+\dfrac{B}{c^2}

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Concept understanding — Mass Energy Equivalence

Mass Energy Equivalence: From Intuition to the Formula

Imagine you have a lump of coal. You know you can burn it to get heat, and that heat can run a steam engine. The energy you get out seems to come from the chemical bonds in the coal. But what if I told you that the coal itself — just sitting there, not burning — already contains a staggering amount of energy locked inside its very mass? That is the core idea of mass-energy equivalence.

The Intuition: Mass is Frozen Energy

Think of mass as a kind of "frozen" or "stored" energy. When you burn coal, you are only releasing a tiny fraction of this stored energy — the energy in the chemical bonds. The rest of the mass remains as matter. But if you could somehow completely convert that lump of coal into pure energy, you would get an unimaginable amount — enough to power a city for years.

This is not a metaphor. Mass and energy are not two separate things that can be converted into each other like dollars and rupees. They are the same fundamental thing, just in different forms. Mass is a highly concentrated form of energy. Energy, when concentrated enough, behaves like mass.

The Precise Statement

The relationship is given by the most famous equation in physics:

E = mc^2

Where:

$E$ is the energy equivalent of the mass (in joules, J)
$m$ is the mass (in kilograms, kg)
$c$ is the speed of light in vacuum ( $3 \times 10^8$ m/s)

The speed of light is a huge number. Squaring it makes it enormous. This is why a tiny amount of mass corresponds to a colossal amount of energy.

What This Equation Actually Means

The equation tells you exactly how much energy is "stored" inside any object with mass $m$ . If you could annihilate that mass completely, you would get $E$ joules of energy.

Example: A 1 kg mass (like a litre of water) contains:

E = 1 \times (3 \times 10^8)^2 = 9 \times 10^{16} \text{ J}

That is 90 quadrillion joules — roughly the energy released by a 20-megaton nuclear bomb. This is not energy you can normally access; it is locked inside the nucleus of atoms.

Where Does This Show Up in Real Life?

Nuclear Reactions: In nuclear fission (splitting atoms) or fusion (joining atoms), a tiny fraction of the mass of the nucleus is converted into energy. The mass of the products is slightly less than the mass of the reactants. The "missing" mass has become energy — exactly as $E=mc^2$ predicts. This is how the Sun works and how nuclear power plants generate electricity.
Particle Physics: When a particle and its antiparticle meet, they annihilate completely into pure energy (usually gamma rays). The energy produced equals $mc^2$ for the two particles. …

Why this formula?

Why $E = mc^2$ — The Reasoning Behind Mass-Energy Equivalence

The formula $E = mc^2$ is not a random guess. Einstein arrived at it by thinking deeply about what happens to energy when you move an object. The core insight: if an object gains energy, it must behave as if it has gained mass.

The Starting Point: Relativistic Momentum

In special relativity, the momentum of a particle is not simply $p = mv$ . Instead, it is:

p = \frac{m_0 v}{\sqrt{1 - v^2/c^2}}

where $m_0$ is the rest mass (mass measured when the object is at rest). This formula already tells us something strange: as speed approaches $c$ , momentum shoots toward infinity — no object with mass can reach the speed of light.

The Energy-Momentum Relation

Einstein then asked: what is the correct expression for kinetic energy that matches this new momentum? In classical physics, kinetic energy is $K = \frac{1}{2}mv^2$ . But that formula fails at high speeds.

The relativistic kinetic energy turns out to be:

K = \frac{m_0 c^2}{\sqrt{1 - v^2/c^2}} - m_0 c^2

This looks odd — why subtract $m_0 c^2$ ? Because when $v = 0$ , the first term becomes $m_0 c^2$ , and we want $K = 0$ at rest. So the subtraction gives zero kinetic energy when the object is stationary.

Note

The term $m_0 c^2$ appears naturally as a rest energy — energy that an object has simply because it has mass, even when completely at rest.

The Crucial Step: What Happens When You Add Energy?

Now consider a box that emits light (photons) in opposite directions. The light carries away energy. Classical physics says the box loses energy but its mass stays the same. Einstein showed this cannot be true.

The argument (simplified): if the box emits a pulse of light with energy $E$ , the light carries momentum $p = E/c$ . By conservation of momentum, the box recoils. But after the light is absorbed by the opposite wall, the box stops. The net effect: the box has moved slightly. Its center of mass has shifted — unless the energy carried by the light also carried mass.

Important

For the center of mass of the entire system (box + light) to remain stationary, the light must behave as if it has an effective mass $m = E/c^2$ . Therefore, energy itself has inertia.

The Full Formula

The total energy of any object — moving or at rest — is:

E = \frac{m_0 c^2}{\sqrt{1 - v^2/c^2}}

For an object at rest ( $v = 0$ ), this reduces to:

E = m_0 c^2

For a moving object, the total energy is the sum of rest energy and kinetic energy:

E = m_0 c^2 + K

E = mc^2

Why It's Not Just a "Conversion" …

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Step 1. A bound nucleus has less mass than its separated constituent nucleons, by an amount equal to the mass-equivalent of its binding energy: this is exactly the mass-defect relation $\Delta m=(Zm_p+Nm_n)-M$ .

Step 2. Rearranging for $M$ : $M=Zm_p+Nm_n-\Delta m$ . Since the binding energy is $B=(\Delta m)c^2$ , the mass defect itself is $\Delta m=B/c^2$ (note: dividing by $c^2$ , not multiplying, since $B$ has units of energy and $\Delta m$ has units of mass).

Step 3. Substituting: $M(N,Z)=NM_n+ZM_p-\dfrac{B}{c^2}$ , which is option (c). …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

CBSE 2026Set A1 markMCQ
$Q. How many joules of energy will be released due to mass defect of 1 mg? (A) 9 \times 10^6 J (B) 3 \times 10^8 J (C) 9 \times 10^10 J (D) 3 \times 10^2 J$
›Reveal solutionSolution

E = mc² with m = 1 mg = 1 × 10⁻⁶ kg gives 9 × 10¹⁰ J.

Using Einstein's mass–energy relation $E = mc^2$ :

Mass defect $m = 1\ \text{mg} = 1\times10^{-6}\ \text{kg}$ .

Speed of light $c = 3\times10^{8}\ \text{m/s}$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The equivalent energy of 10 g of substance is$
(a) 9 x 10^14 J
(b) 9 x 10^13 J
(c) 3 x 10^16 J
(d) 6 x 10^13 J
›Reveal solutionSolution

Mass-energy equivalence E = m*c^2 converts a mass of 10 g directly into an energy of 9 x 10^14 J.

Einstein's mass-energy equivalence relation states that mass and energy are interconvertible, related by

E = m * c^2

where m is the mass (in kg) and c = 3 x 10^8 m/s is the speed of light in vacuum. Here m = 10 g = 0.01 kg. Substituting:

E = 0.01 * (3 x 10^8)^2 = 0.01 * 9 x 10^16 = 9 x 10^14 J

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The energy equivalent of 1 mg of substance is$
(a) 9 × 10^10 J
(b) 9 × 10^13 J
(c) 1.35 × 10^14 J
(d) 3 × 10^13 J
›Reveal solutionSolution

Einstein's mass–energy relation $E=mc^2$ shows even a tiny mass carries an enormous amount of energy.

Given: $m = 1\ \text{mg} = 1\times10^{-6}\ \text{kg}$ , $c = 3\times10^8\ \text{m/s}$ .

$E = mc^2 = (1\times10^{-6})(3\times10^8)^2 = (1\times10^{-6})(9\times10^{16})$

$E = 9\times10^{10}\ \text{J}$

…

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CBSE 2025Set IMPROVEMENT1 mark
$Q. Write the energy equivalent of one atomic mass unit in joules.$
›Reveal solutionSolution

Using Einstein's mass–energy relation $E = mc^2$ with $m = 1\,\text{u} = 1.6605\times10^{-27}\,\text{kg}$ gives the energy equivalent of 1 amu.

The energy equivalent of a mass $m$ is $E = mc^2$ . For one atomic mass unit, $m = 1.6605\times10^{-27}\,\text{kg}$ and $c = 3\times10^{8}\,\text{m/s}$ :

$E = (1.6605\times10^{-27})\times(3\times10^{8})^2 = 1.6605\times10^{-27}\times 9\times10^{16}$ …

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CBSE 2025Set D1 markMCQ
$Q. Equivalent energy of 1 amu is (A) 190 MeV (B) 139 MeV (C) 913 MeV (D) 931 MeV$
›Reveal solutionSolution

Mass–energy equivalence gives 1 u ≈ 931 MeV.

Using Einstein's relation E = mc² with 1 u = 1.6605×10⁻²⁷ kg:

$E = (1.6605\times10^{-27})(3\times10^{8})^{2} = 1.4924\times10^{-10}\ \text{J}$

Converting to MeV (1 MeV = 1.602×10⁻¹³ J): …

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CBSE 2024Set A1 mark
$Q. Match Column 'A' with Column 'B' and write the correct pair. Column 'A' item: 'Mass-energy equivalence relation'. Column 'B' options:$
(i) De-Broglie
(ii) Maxwell
(iii) Ohm
(iv) Einstein
(v) Coulomb
(vi) Lenz
(vii) Young.
›Reveal solutionSolution

Mass-energy equivalence, E = mc², was given by Albert Einstein.

Albert Einstein, through his theory of relativity, showed that mass and energy are equivalent and interconvertible, related by:

$E = mc^2$ …

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CBSE 2024Set ANNUAL1 mark
$Q. Write Einstein's mass-energy equivalent relation.$
›Reveal solutionSolution

Einstein's mass-energy equivalence states that mass itself is a form of energy, related by the constant c².

Einstein's special theory of relativity showed that mass and energy are two forms of the same physical quantity and are interconvertible. The mass-energy equivalence relation is:

$E = mc^2$

…

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CBSE 2023Set ANNUAL1 mark
$Q. Answer in one word/sentence: Give the mass-energy equivalence equation.$
›Reveal solutionSolution

Einstein's equation E = m*c^2 shows that mass and energy are equivalent and interconvertible, with c^2 (a very large number) as the conversion factor.

According to Einstein's theory, any mass m has an associated rest energy E given by:

E = m*c^2

…

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CBSE 2022Set ANNUAL1 mark
$Q. Calculate the energy equivalent of 1 gm. of substance.$
›Reveal solutionSolution

Use $E = mc^2$ with $m = 1\,\text{g} = 10^{-3}\,\text{kg}$ .

By Einstein's mass-energy equivalence relation, $E = mc^2$ .

Here $m = 1\,\text{gm} = 1\times10^{-3}\,\text{kg}$ and $c = 3\times10^{8}\,\text{m/s}$ (given constant).

…

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CBSE 2022Set I1 markMCQ
$Q. Which of the following relations is correct for mass and energy? (A) m = E (B) m^2 = E (C) mc^2 = E (D) m = \sqrtE / 2$
›Reveal solutionSolution

Mass and energy are equivalent: E = mc².

Einstein's special theory of relativity established that mass and energy are interchangeable. A mass m is equivalent to an energy:

$E = mc^2$

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. Match Column A item 'Energy - mass equality' with the correct item in Column B.$
(a) kg m^2
(b) Poise
(c) Cp - Cv = R
(d) E = mc^2
(e) 1/frequency
(f) distance
(g) 24 hours
›Reveal solutionSolution

The mass-energy equivalence relation, discovered by Einstein, is E = mc^2, where c is the speed of light in vacuum.

This relation states that a mass m has an equivalent rest energy E = mc^2, and conversely that energy has an equivalent mass. It underlies nuclear energy release (a small mass defect converts into a large a …

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CBSE 2019Set HE2341 markMCQ
$Q. The energy equivalent of 1 gram of substance is:$
(i) 9×10^6 Joule
(ii) 3×10^13 Joule
(iii) 3×10^6 Joule
(iv) 9×10^13 Joule
›Reveal solutionSolution

E = mc² with m = 1 g = 10⁻³ kg and c = 3×10⁸ m/s gives E = 9×10¹³ J.

Einstein's mass-energy equivalence relation states that mass and energy are interconvertible:

$E = mc^2$

Here $m = 1\text{ g} = 1\times10^{-3}\text{ kg}$ and $c = 3\times10^{8}\text{ m/s}$ (speed of light).

$E = (1\times10^{-3})\times(3\times10^{8})^2 = (1\times10^{-3})\times(9\times10^{16}) = 9\times10^{13}\ \text{J}$ …

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Q.MpM_pMp​ denotes the mass of a proton and MnM_nMn​ denotes the mass of a neutron. A given nucleus of binding energy BBB contains ZZZ protons and NNN neutrons. The mass M(N,Z)M(N,Z)M(N,Z) of the nucleus is given by (where ccc is the speed of light)

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Q. $M_p$ denotes the mass of a proton and $M_n$ denotes the mass of a neutron. A given nucleus of binding energy $B$ contains $Z$ protons and $N$ neutrons. The mass $M(N,Z)$ of the nucleus is given by (where $c$ is the speed of light)

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