Q.The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about $10^{-40}$. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.

The key idea here is that the electron in the first orbit of hydrogen moves in a circular path. Its frequency is simply how many times it goes around the circle per second — that is, the orbital speed divided by the circumference of the orbit. This is a direct application of the relation between speed, radius, and frequency for uniform circular motion.

1. Recall the relationship between speed, radius, and frequency For an object moving in a circle of radius $r$ with constant speed $v$, the time for one complete revolution (the period $T$) is the distance around the circle divided by the speed:

The frequency $f$ is the reciprocal of the period:

2. Plug in the given values We are told:
   • Orbital radius $r = 0.5 \times 10^{-10} \, \text{m}$
   • Orbital velocity $v = 2.2 \times 10^6 \, \text{m/s}$ So:

3. Simplify step by step First, compute the denominator:

(since $2 \times 0.5 = 1$).

So:

4. Evaluate numerically Using $\pi \approx 3.1416$:

Thus:

5. Match with the options The closest value is $6.98 \times 10^{15} \, \text{Hz}$, which is option (C). The small difference is due to rounding of $\pi$ and the given numbers.

For the electron in Bohr's first orbit, the orbital speed follows from combining the Coulomb–centripetal balance with angular-momentum quantization ($mvr=\tfrac{h}{2\pi}$):

The frequency of revolution is the speed divided by the circumference of the orbit:

Concept & Intuition

When an electron moves perpendicular to a uniform magnetic field, it experiences a centripetal Lorentz force and follows a circular path. In quantum mechanics, the angular momentum of such an orbit is quantized in units of $\hbar = h/(2\pi)$. This is analogous to Bohr’s quantization for atomic orbits, but here the centripetal force is magnetic rather than electrostatic. The kinetic energy is purely the energy of circular motion, and we can find it by combining the force balance with the quantization condition.

Step-by-step reasoning

1. Force balance for circular motion For an electron of mass $m$ and charge $e$ moving with speed $v$ in a circle of radius $r$ perpendicular to a uniform magnetic field $B$, the magnetic (Lorentz) force provides the centripetal force:

Cancelling one $v$ gives:

2. Quantization of angular momentum Bohr’s postulate for a stationary orbit states that the angular momentum is an integer multiple of $\hbar = h/(2\pi)$:

Substitute $v$ from step 1 into this:

Hence the radius is:

3. Kinetic energy expression Kinetic energy is $K = \frac{1}{2} m v^2$. Using $v = e B r / m$ from step 1:

Now substitute $r^2$ from step 2:

4. Match with options The result $\frac{n h e B}{4 \pi m}$ is exactly option (B).

The Bohr model of the atom was a revolutionary step in explaining why atoms emit only specific frequencies of light. The key insight is that classical physics would allow an electron to spiral into the nucleus, emitting a continuous spectrum. Bohr fixed this by imposing a quantisation condition: the electron’s angular momentum can only take certain discrete values. This single assumption leads to fixed orbits, discrete energy levels, and thus a line spectrum — not a continuous one.

Let’s examine each option carefully.

1. Option A: “Predicts continuous emission spectra for all atoms”

   This is false. The Bohr model explicitly predicts discrete (line) spectra, not continuous. A continuous spectrum would come from a free electron radiating energy as it spirals in — exactly what Bohr’s quantisation prevents. So A is wrong.

2. Option B: “Assumes that the angular momentum of electrons is quantised”

   This is the core postulate of the Bohr model. Bohr stated that the electron’s angular momentum $L$ must be an integer multiple of $\hbar = h/(2\pi)$:

This quantisation is what gives rise to stable orbits and discrete energy levels. So B is correct.

3. Option C: “Predicts same emission spectrum for singly ionised neon atom and hydrogen atom”

   Singly ionised neon (Ne⁺) has 9 electrons, while hydrogen has 1. The Bohr model works well only for one-electron systems (like H, He⁺, Li²⁺). For Ne⁺, the remaining electrons screen the nucleus and interact with each other, so the simple Bohr formula fails. Even if we considered a one-electron ion, the nuclear charge $Z$ differs: for H, $Z=1$; for Ne⁺, $Z=10$. The energy levels scale as $Z^2$, so the spectra are completely different. Thus C is false.

4. Option D: “Predicts same emission spectrum for singly ionised neon atom and singly ionised helium atom”

   Singly ionised helium (He⁺) is a one-electron ion with $Z=2$. Singly ionised neon (Ne⁺) is not a one-electron system (it has 9 electrons). Even if we ignored that, the nuclear charges are different ($Z=2$ vs $Z=10$), so the spectra would not match. So D is false.

The key idea is that Bohr’s model quantizes angular momentum:

and the Coulomb force provides the centripetal force:

From these two equations, we can solve for $v$, $r$, and the total energy $E$ in terms of $n$. The dependencies are not arbitrary — they follow directly from combining these relations.

1. Find how radius $r$ depends on $n$. From the angular momentum condition: $v = \frac{n h}{2\pi m r}$. Substitute into the force equation:

Simplify:

Multiply both sides by $r^3$:

So

Thus radius $\propto n^2$.

2. Find how speed $v$ depends on $n$. From angular momentum: $v = \frac{n h}{2\pi m r}$. Since $r \propto n^2$, we get

So speed $\propto 1/n$.

3. Find how total energy $E$ depends on $n$. Total energy = kinetic + potential:

Using the force equation $\frac{k e^2}{r^2} = \frac{m v^2}{r}$, we have $m v^2 = \frac{k e^2}{r}$.

Then kinetic energy = $\frac{1}{2} m v^2 = \frac{k e^2}{2r}$.

So

Since $r \propto n^2$,

The magnitude (or the absolute value) varies as $1/n^2$, so energy $\propto 1/n^2$.

4. Match with the options. Speed $\propto 1/n$, energy $\propto 1/n^2$, radius $\propto n^2$. That is exactly option (B).

The problem is about how the size of an atom changes when the orbiting particle’s mass changes. In the Bohr model, the radius of an orbit depends on the reduced mass of the two-body system (nucleus + orbiting particle). Most students forget that the electron’s mass appears in the denominator of the radius formula — so a heavier particle actually orbits closer to the nucleus. Let’s walk through it carefully.

1. Recall the Bohr radius formula for a hydrogen-like atom For a nucleus of charge $+Ze$ and an orbiting particle of mass $m$ and charge $-e$, the radius of the $n$-th orbit is:

But this formula assumes the nucleus is infinitely heavy. In reality, the electron and nucleus both orbit their common centre of mass, so we must use the reduced mass $\mu$:

where $m$ is the orbiting particle’s mass and $M$ is the nucleus mass. For hydrogen, $M \approx 1836 m_e$, so $\mu \approx m_e$. The correct radius is:

So the radius is inversely proportional to the reduced mass.

2. What changes in the problem?
   • Normal hydrogen: orbiting particle is an electron of mass $m_e$, nucleus is a proton of mass $m_p$. Reduced mass $\mu_H \approx m_e$ (since $m_p \gg m_e$).
   • Modified atom: orbiting particle has mass $40 m_e$ (same charge $-e$), nucleus is still a proton. The new reduced mass is:

 Since $m_p \approx 1836 m_e$, the denominator is dominated by $m_p$, so:

 More precisely, $\mu'$ is very close to $40 m_e$ because the proton is still much heavier than the new particle. So the reduced mass increases by a factor of about 40.

3. Effect on the ground state radius

For the modified atom in its ground state ($n=1$):

Compared to the normal hydrogen ground state radius $r_1 \propto \frac{1}{m_e}$, we get:

So the heavy particle orbits 40 times closer to the nucleus.

4. First excited state of normal hydrogen The first excited state corresponds to $n=2$. For normal hydrogen:

because radius scales as $n^2$.

5. Compute the required ratio We want:

So the ratio is $160 : 1$.

The concept here is angular momentum of a particle in circular motion. For a point mass moving in a circle, the magnitude of its orbital angular momentum about the center is simply the product of its linear momentum ($mv$) and the radius ($r$), because the velocity is perpendicular to the radius vector. This is a direct application of $L = m v r$, not requiring integration or calculus.

1. Identify the formula: For a particle of mass $m$ moving with speed $v$ in a circle of radius $r$, the angular momentum about the center is

This works because the angle between the velocity and the radius is $90^\circ$, so $\sin 90^\circ = 1$.

2. Plug in the given values:

3. Multiply step by step (keeping track of powers of 10): First, multiply the coefficients:

(More precisely: $20.02 \times 0.5 = 10.01$, plus $20.02 \times 0.029 \approx 0.5806$, sum $\approx 10.5906$.)

4. Combine the powers of ten:

So the product is approximately

5. Round to three significant figures (since all inputs have two or three significant figures):

Step 1 — The Bohr energy formula for a hydrogen-like species

For any one-electron (hydrogen-like) atom or ion with nuclear charge $Z$:

The $Z^2$ arises because a larger nuclear charge binds the electron more tightly (the Coulomb attraction scales with $Z$, and the orbit radius shrinks as $1/Z$); the $1/n^2$ is the usual orbit-quantisation result. $He^+$ has one electron and $Z = 2$, so the formula applies to it.

Step 2 — Hydrogen, second orbit

$Z = 1$, $n = 2$:

Step 3 — $He^+$, third orbit

$Z = 2$, $n = 3$:

Step 4 — Express it in terms of $E_2$ (take the ratio)

Numerically: $\frac{16}{9}\times(-3.4) = -6.04$ eV. ✓ Consistent.

Step 5 — Sanity check on the sign and magnitude

Both energies are negative (bound states). The $He^+$ level is more negative (more tightly bound) than H's $n=2$ level, so the multiplying factor must be greater than 1 — which $\frac{16}{9} = 1.78$ is. This immediately kills options (A) $\frac{9}{16}$ and (C) $\frac{3}{16}$ (both $< 1$), and (D) $\frac{16}{3} = 5.33$ would give $-18.1$ eV, far too deep for an $n=3$ level of $He^+$.

ANSWER: B

Step 1 — Bohr's radius and speed.

For hydrogen,

(The $n^2$ growth of the orbit and the $1/n$ slowing of the electron both come from the quantisation condition $mvr=\frac{nh}{2\pi}$ combined with the Coulomb force providing the centripetal force.)

Step 2 — Period of revolution.

So $T_n\propto n^{3}$ — the electron in a higher orbit takes dramatically longer to go round, because the orbit is much bigger and it moves more slowly.

Step 3 — Apply to the given states.

Ground state: $n=1\Rightarrow T_1=T$.

First excited state: $n=2$ (not $n=1$ — this is the usual trap).

Step 4 — Distractors.

$2T$ would follow from $T\propto n$, $4T$ from $T\propto n^{2}$ (a common error, using only the radius). The correct cubic dependence gives $8T$.

ANSWER: D