Q.The gravitational force between a H-atom and another particle of mass $m$ will be given by Newton's law: $F = G\dfrac{M\,m}{r^2}$, where $r$ is in km and

The key idea is that the gravitational mass of a bound system includes the mass equivalent of its binding energy. For a hydrogen atom, the total mass is the sum of the proton and electron masses minus the mass equivalent of the binding energy.

Reasoning:

1. Newton's law uses the gravitational mass $M$ of the hydrogen atom. In general relativity, the source of gravity is the total energy content, including rest masses and binding energy.
2. The binding energy $B = 13.6\ \text{eV}$ is the energy required to separate the electron from the proton. The system's total energy is $(m_p + m_e)c^2 - B$, so its effective mass is $M = m_p + m_e - B/c^2$.
3. The potential energy $|V|$ is not the binding energy — for the hydrogen atom, $|V| = 27.2\ \text{eV}$, while the binding energy $B = 13.6\ \text{eV}$ (the difference is the kinetic energy). The correct reduction is by $B/c^2$, not $|V|/c^2$.

The question tests a subtle but beautiful point: when Newton’s law says $F = G M m / r^2$, the $M$ is the gravitational mass of the hydrogen atom. In Einstein’s relativity, gravitational mass is equivalent to total energy (including rest energy) divided by $c^2$. So we must account for all contributions to the atom’s energy — not just the masses of its constituents, but also their kinetic and potential energies, and the binding energy that holds them together.

Let’s walk through it.

1. Rest masses alone are not enough. A hydrogen atom consists of a proton and an electron. If we simply added their rest masses, we’d get $m_p + m_e$. But the atom is a bound system: the electron is in a quantum state around the proton, with kinetic energy $K$ and negative potential energy $V$ (taking $V=0$ at infinity). The total energy of the atom is

For the ground state, the binding energy $B = 13.6\ \text{eV}$ is defined as the energy needed to separate the atom into a free proton and a free electron at rest. That means

So $K+V = -B$.

2. The gravitational mass comes from total energy. By Einstein’s equivalence principle, the gravitational mass $M$ of any object is its total energy divided by $c^2$:

Substituting $K+V = -B$ gives

3. What about the potential energy magnitude $|V|$?

   Option (D) uses $|V|$ instead of $B$. But $|V|$ alone is not the binding energy — the kinetic energy also contributes. For the hydrogen ground state, the virial theorem tells us $K = -\frac{1}{2}V$, so $B = -(K+V) = -\left(-\frac{1}{2}V + V\right) = -\frac{1}{2}V$. That means $|V| = 2B$, which is not the correct correction. So (D) is wrong.

4. Why not (A) or (C)?

   Option (A) ignores the binding energy entirely — it would be correct only if the atom were just a loose collection of two particles. Option (C) is nonsense: the gravitational mass is certainly related to the atom’s mass.

Method: Computing the Effective (Gravitational) Mass of a Bound System

Use this whenever a problem asks for the mass that should appear in a gravity/momentum formula for a bound system of particles (e.g. an atom, nucleus, or molecule), rather than just adding up the constituents' rest masses.

Steps

Step 1: Write the total rest energy of the free (unbound) constituents

If the system is made of particles of mass $m_1, m_2, \dots$, their combined rest energy before binding is

Step 2: Subtract the binding energy

Binding energy $B$ is, by definition, the energy that had to be removed to pull the system together (equivalently, the energy needed to pull it back apart). So the system's actual total energy is less than the free-particle sum:

Step 3: Convert energy back to an effective mass via $E = Mc^2$

Since mass and energy are equivalent, divide by $c^2$ to get the system's true (gravitational/inertial) mass:

This is the mass that should be substituted into Newton's law of gravitation or any other formula that needs the system's real mass — never the naive sum of constituent rest masses.

Step 4: Don't confuse potential energy magnitude with binding energy

A common trap is substituting the magnitude of the potential energy $|V|$ in place of $B$. They are only equal when there's no kinetic energy contribution. In general, $B = -(K+V)$, so you need the virial theorem or the system's actual total mechanical energy to relate $|V|$ to $B$ correctly — don't assume they're interchangeable without checking.