Q.The deuteron is bound by nuclear forces just as H-atom is made up of p and e bound by electrostatic forces. If we consider the force between neutron and proton in deuteron as given in the form of a Coulomb potential but with an effective charge $e'$: $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{e'^2}{r}$, estimate the value of $(e'/e)$ given that the binding energy of a deuteron is $2.2\ \text{MeV}$.

Concept & Intuition

When an alpha particle (charge $2e$) is fired head-on toward a heavy nucleus (charge $Ze$), it slows down as it climbs the Coulomb potential hill. At the point of closest approach, its kinetic energy has been completely converted into electrostatic potential energy — it momentarily stops before being repelled back. This is a pure energy conservation problem:

Solving for $r_{\text{min}}$ gives the distance of closest approach. The trick is to use consistent units — nuclear physicists often work in MeV and femtometers, using the handy constant $e^2/(4\pi\varepsilon_0) \approx 1.44\ \text{MeV·fm}$.

Step-by-step solution

1. Write the energy conservation equation At the turning point, all kinetic energy $K$ becomes Coulomb potential energy:

Here $Z = 50$ (atomic number of the target nucleus), and the alpha particle has charge $2e$.

2. Solve for $r_{\text{min}}$

3. Use the convenient constant The product $\frac{e^2}{4\pi\varepsilon_0} \approx 1.44\ \text{MeV·fm}$. So:

4. Plug in numbers $Z = 50$, $K = 6.5\ \text{MeV}$:

5. Convert to the units used in the options

   The options are given in fm (femtometers) but note that (C) is 0.0221 fm — that’s suspiciously small. Actually, check: 22.15 fm = $22.15 \times 10^{-15}$ m. But 1 picometer (pm) = $10^{-12}$ m = 1000 fm. So 22.15 fm = 0.02215 pm. The options list (C) as 0.0221 fm — but that would be 0.0221 fm, which is 1000 times smaller. Wait — let’s re-read the options carefully:

   (A) 0.221 fm

   (B) 1.101 fm

   (C) 0.0221 fm

   (D) 4.42 fm …

Concept and intuition:

When an alpha particle (charge $+2e$) is fired head-on at a gold nucleus (charge $+79e$), it slows down as it approaches because of the repulsive electrostatic force. At the point of closest approach, its speed becomes zero — all its initial kinetic energy has been converted into electrostatic potential energy. So we can equate the kinetic energy $K$ to the potential energy $U$ at that distance $r$:

This is a direct application of energy conservation in a purely electrostatic field.

Step-by-step solution:

1. Identify the charges and constants

   Alpha particle charge: $q_1 = 2e = 2 \times 1.6 \times 10^{-19} \, \mathrm{C}$

   Gold nucleus charge: $q_2 = 79e = 79 \times 1.6 \times 10^{-19} \, \mathrm{C}$

   Distance of closest approach: $r = 10 \times 10^{-14} \, \mathrm{m} = 10^{-13} \, \mathrm{m}$

   Coulomb constant: $k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \, \mathrm{N \cdot m^2/C^2}$

2. Write the potential energy formula

3. Substitute the values

4. Simplify step by step First, compute the product of charges:

$2 \times 79 = 158$ and $(1.6)^2 = 2.56$, so:

Thus numerator = $404.48 \times 10^{-38} \, \mathrm{C^2}$

Now divide by $r = 10^{-13} \, \mathrm{m}$:

At the distance of closest approach $r$, all kinetic energy $E$ becomes electrostatic potential energy:

$E=\frac{1}{4\pi\varepsilon_0}\frac{q_\alpha q_{Au}}{r}\implies r\propto\frac{1}{E}.$ …

The key idea is that Bohr’s model isn’t limited to electrons. It says angular momentum in any bound orbital motion comes in integer multiples of $\frac{h}{2\pi}$ (i.e., $\hbar$). For Earth going around the Sun, we can compute its classical angular momentum and then see which integer $n$ it corresponds to.

1. Write Bohr’s quantization condition Bohr’s postulate:

Here $m$ is Earth’s mass, $v$ its orbital speed, $r$ the orbit radius, $h$ Planck’s constant, and $n$ the quantum number we want.

2. Plug in the given numbers

   • $m = 6 \times 10^{24} \ \text{kg}$
   • $v = 3 \times 10^{4} \ \text{m/s}$
   • $r = 1.5 \times 10^{11} \ \text{m}$
   • $h = 6.63 \times 10^{-34} \ \text{J·s}$ (standard value)

   First compute the classical angular momentum:

Multiply stepwise:

Powers of ten: $10^{24} \times 10^{4} \times 10^{11} = 10^{39}$

So $L = 27 \times 10^{39} = 2.7 \times 10^{40} \ \text{kg·m}^2/\text{s}$

3. Find $n$ From $L = n \hbar$, we have

Compute $\hbar$: …

Step 1 — The Bohr radius formula

In Bohr's model of the hydrogen-like atom, the radius of the $n$-th permitted orbit is

where $a_0$ is the radius of the first orbit ($n=1$), called the Bohr radius. Here $a_0 = 0.529~\overset{\circ}{A}$ is given directly (its theoretical value $a_0 = \dfrac{4\pi\varepsilon_0\hbar^2}{m_e e^2} \approx 0.529~\overset{\circ}{A}$ is exactly this number, so no separate derivation of $a_0$ is needed).

Step 2 — Apply for the third orbit

For the third orbit, $n = 3$, so $n^2 = 9$:

Step 3 — Check the other options …

Step 1 — Find the orbit number $n$.

Bohr's second postulate quantises angular momentum in units of $h/2\pi$:

Given $L=\dfrac{3h}{2\pi}$, comparing gives

Step 2 — Relate kinetic energy to total energy.

For an electron bound by the Coulomb force, the electrostatic attraction supplies the centripetal force:

Hence the total energy $E=K+U=-K$, i.e. $K=-E$ (the virial theorem for an inverse-square field).

Step 3 — Put in the hydrogen energy level. …

Step 1 — The Bohr energy levels of hydrogen.

Bohr's model gives the total energy (kinetic + electrostatic potential) of the electron in the $n$-th orbit as

The energy is negative because the electron is bound: work must be done to pull it to infinity, where $E=0$.

Step 2 — Substitute $n = 2$ (the second orbit).

Step 3 — Check the distractors. …