Q.A $100\ \text{W}$ sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is $589\ \text{nm}$.

Step 1 — Bohr's radius formula

For a hydrogen-like atom, Bohr's quantisation of angular momentum ($mvr = n\hbar$) combined with the Coulomb-force–centripetal-force balance gives

For hydrogen ($Z=1$), $r_n = n^2 a_0$ with $a_0 = 0.529\ \text{Å}$.

Step 2 — From radius to area

The orbit is a circle, so the area enclosed is

Since $r_n \propto n^2$,

This $n^4$ (not $n^2$) is the crux of the question — the squaring of an already-squared quantity.

Step 3 — Identify the two states

• Ground state: $n=1$.
• First excited state: the next level up, $n=2$. (Not $n=3$ — the "first excited" state is the first level above the ground state.)

Step 4 — Take the ratio …

Step-by-step reasoning

1. Convert the energy to joules.

2. Solve for wavelength.

3. Convert to convenient units.

This matches the option listing 900 Å. …

1. The Bohr radius law

Quantising the angular momentum ($mvr = nh/2\pi$) and balancing the Coulomb force against the centripetal requirement gives, for a hydrogen-like atom,

where $a_0 = 0.53\ \text{Å}$ is the Bohr radius. For hydrogen $Z = 1$, so simply

The radius grows as the square of the principal quantum number — that quadratic dependence is the whole content of the problem.

2. Set up the ratio

Ground state: $r_1 = 0.53\ \text{Å}$ (with $n=1$, consistent with the formula).

Final state after the collision: $r_n = 2.12\ \text{Å}$.

Taking the ratio kills $a_0$ entirely:

3. Solve

4. Physical reading …

The key here is to connect the frequency of revolution — how many times per second the electron circles the nucleus — to the orbital radius and velocity. In Bohr's model, the electron moves in a circular orbit under electrostatic attraction, and its angular momentum is quantized. Frequency is simply $v/(2\pi r)$, so if we find how $v$ and $r$ depend on $n$, we can combine them.

Let's work through it step by step.

1. Write the force balance for a stable orbit. The centripetal force is provided by the Coulomb attraction between the electron and the proton:

where $m$ is the electron mass, $v$ its speed, $r$ the orbit radius, $k = 1/(4\pi\epsilon_0)$, and $e$ the elementary charge.

2. Apply Bohr's quantization condition. Angular momentum is an integer multiple of $\hbar = h/(2\pi)$:

3. Solve for $r$ in terms of $n$. From the quantization condition, $v = n\hbar/(mr)$. Substitute into the force equation:

Simplify:

Multiply both sides by $r^3$:

So:

This shows $r \propto n^2$.

4. Find $v$ in terms of $n$. From $mvr = n\hbar$, we have $v = n\hbar/(mr)$. Substitute $r \propto n^2$:

So $v \propto 1/n$.

5. Now compute the frequency of revolution. Frequency $f$ is the number of orbits per second: $f = \frac{v}{2\pi r}$ …