Q.In a nuclear reactor, moderators slow down the neutrons which come out in a fission process. The moderator used have light nuclei. Heavy nuclei will not serve the purpose because

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

   Our computed value is ~22 fm, which is not among these. This suggests the options might actually be in picometers (pm) but mislabeled? Let’s check: 22 fm = 0.022 pm. That matches 0.0221 — so option (C) is almost certainly meant to be 0.0221 pm (or equivalently 22.1 fm). In many textbooks, the distance of closest approach for such parameters is indeed about 22 fm.

6. Final check Using exact values:

Rounded to three significant figures: 22.2 fm = 0.0222 pm. Option (C) says 0.0221 fm — but that’s a factor of 1000 off unless the unit is pm. Given typical exam conventions, (C) is the intended answer.

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

Multiply by $k = 9 \times 10^9$:

Which is $3.64032 \times 10^{-13} \, \mathrm{J}$.

5. Round to match the options This is $3.64 \times 10^{-13} \, \mathrm{J}$.

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

Same charges for both alphas and the same gold nucleus, so

Hence $\alpha_2:\alpha_1=1:3$.

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 $\hbar$:

Then

Rounding gives $2.57 \times 10^{74}$.

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

• (B) $5125~\text{nm}$ and (C) $4234~\text{nm}$ are both off by orders of magnitude ($1~\overset{\circ}{A} = 0.1~\text{nm}$, so a Bohr-orbit radius here should be a few $\overset{\circ}{A}$ — a few tenths of a nanometre — not thousands of nanometres); these are unit-scale distractors.
• (D) $4.496~\overset{\circ}{A}$ does not equal $9 \times 0.529$ and does not follow from the $n^2$ scaling law.

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 4 — Kinetic energy.

Note the traps: $3.4\,\text{eV}$ and $6.8\,\text{eV}$ correspond to $n=2$ (they are $|E_2|$ and $2|E_2|$), and would be picked by anyone who mis-reads $n$ or confuses $K$ with $|U|$.

ANSWER: B

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.

• $-13.6$ eV is $E_1$ — the ground state ($n=1$), not the second orbit.
• $-1.51$ eV is $E_3=-13.6/9$ — the third orbit.
• Zero is the energy of a free electron at $n\to\infty$ (the ionised atom). Only $-3.4$ eV corresponds to $n=2$.

ANSWER: C