Obtain the first Bohr's radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon ($\mu^-$) of mass about $207m_e$ orbits around a proton].

Obtain the first Bohr's radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a

Additional Exercises · 12.17

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Concept understanding — Bohr Model Quantization

Why does an electron not spiral into the nucleus?

Imagine you're pushing a child on a swing. If you push at random moments, the swing jerks and slows down. But if you push exactly in rhythm with the swing's natural motion, each push adds energy smoothly and the swing goes higher and higher. The swing "prefers" to move at specific frequencies — its natural modes.

An electron orbiting a nucleus is similar, but with a crucial twist from the quantum world. Classical physics says an accelerating charge (like an electron going in a circle) must continuously radiate energy. If that were true, the electron would lose energy, spiral into the nucleus, and atoms would collapse in a flash of light. But atoms are stable. So something is fundamentally different.

The radical idea: allowed orbits only

Niels Bohr proposed in 1913 that the electron cannot occupy just any orbit. It can only exist in certain stationary states — orbits where it does not radiate energy. These are like the swing's natural frequencies, but for an electron.

The key condition that picks out these special orbits is called quantization of angular momentum.

L = n \frac{h}{2\pi}, \quad n = 1, 2, 3, \dots

Here $L$ is the orbital angular momentum of the electron, $h$ is Planck's constant, and $n$ is a positive integer called the principal quantum number.

What this means physically

Angular momentum for a circular orbit is $L = m v r$ , where $m$ is the electron mass, $v$ its speed, and $r$ the orbit radius. So the quantization condition becomes:

m v r = n \frac{h}{2\pi}

This is not a formula you derive — it is a postulate, a rule that nature follows. Bohr had no deeper explanation for why this rule works; he simply noticed it gave the right answers for hydrogen's spectrum.

Note

The quantity $\frac{h}{2\pi}$ appears so often that it has its own symbol: $\hbar$ (h-bar). So the condition is often written as $L = n\hbar$ .

What it predicts

Combining this quantization with Newton's law for circular motion (centripetal force = Coulomb attraction) gives:

Radius of the $n$ th orbit: $r_n = n^2 a_0$ , where $a_0 = 0.529 \, \text{Å}$ is the Bohr radius (the smallest orbit, $n=1$ ).
Energy of the $n$ th orbit: $E_n = -\frac{13.6 \, \text{eV}}{n^2}$

The negative sign means the electron is bound to the nucleus. As $n$ increases, the orbit gets larger and the energy becomes less negative (closer to zero).

The key insight for exams

Bohr quantization is not a derivation — it is a condition you apply. When you see a problem about hydrogen-like atoms (one electron), you:

Write $mvr = n\hbar$
Write the force balance: $\frac{mv^2}{r} = \frac{kZe^2}{r^2}$ (for nuclear charge $Ze$ )
Solve for $r$ and $v$ in terms of $n$ …

Why this formula?

Why Angular Momentum is Quantised in the Bohr Model

The Bohr model's most famous result — that angular momentum comes only in integer multiples of $\frac{h}{2\pi}$ — is not an arbitrary assumption. It follows directly from a single, elegant idea: the electron's wave must close on itself.

The core problem Bohr faced

By 1913, physicists knew two things that seemed contradictory:

Rutherford's nuclear model showed electrons orbiting the nucleus.
Maxwell's equations predicted that any accelerating charge (like an orbiting electron) must radiate energy, spiral inward, and collapse in about $10^{-11}$ seconds.

Atoms are stable. Something was missing.

Bohr's breakthrough was to combine the newly discovered quantum idea (Planck's constant $h$ ) with classical mechanics, but only for allowed orbits. The key constraint came from thinking of the electron not as a tiny planet, but as a standing wave.

The de Broglie wavelength argument (the cleanest derivation)

A few years after Bohr, de Broglie proposed that every moving particle has a wavelength:

\lambda = \frac{h}{p} = \frac{h}{mv}

For an electron in a circular orbit of radius $r$ , the circumference is $2\pi r$ . For the wave to be stable — not cancelling itself out — the circumference must contain an integer number of wavelengths:

2\pi r = n\lambda, \quad n = 1, 2, 3, \dots

Substitute $\lambda = h/(mv)$ :

2\pi r = n \cdot \frac{h}{mv}

Rearrange:

mvr = n \cdot \frac{h}{2\pi}

That's it. The left side $mvr$ is the angular momentum $L$ . So:

L = n\hbar, \quad \text{where } \hbar = \frac{h}{2\pi}

Note

This is not a separate postulate — it is a consequence of requiring the electron wave to be a standing wave. If the wave doesn't close on itself, it interferes destructively and the orbit cannot exist.

Why this fixes the energy levels

Once angular momentum is quantised, the rest follows from classical physics. For a circular orbit, the centripetal force is provided by the Coulomb attraction:

\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}

Combine this with $mvr = n\hbar$ and solve for $r$ and $E$ :

r_n = \frac{4\pi\epsilon_0 \hbar^2}{m e^2} \cdot n^2 = a_0 n^2

$E_n = -\frac{m e^4}{8\epsilon_0^2 h^2} \cdot \frac{1}{n^2} = -\frac{13.6 \text{ eV}}{n^2}$ …

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Since the Bohr radius scales as $1/m$ and the ground-state energy scales as $m$ , replacing the electron ( $m_e$ ) with a muon ( $207\,m_e$ ) in the Bohr formulas gives $r_1^\mu \approx 2.56\times10^{-13}\ \text{m}$ (207 times smaller than $a_0$ ) and $E_1^\mu \approx -2.8\ \text{keV}$ (207 times deeper than $-13.6$ eV).

Step 1 -- How the Bohr formulas depend on the orbiting particle's mass.

The Bohr radius and energy levels, derived from balancing the Coulomb force against centripetal force together with angular-momentum quantisation, are

r_n = \frac{n^2\varepsilon_0h^2}{\pi me^2}, \qquad E_n = -\frac{me^4}{8\varepsilon_0^2h^2n^2}

where $m$ is the mass of the orbiting particle (the proton is taken as essentially fixed, since it's far heavier than either an electron or a muon). Notice $r_n \propto 1/m$ while $E_n \propto m$ -- heavier orbiting particles sit in smaller, more tightly bound orbits.

Step 2 -- Substitute the muon's mass.

A muonic hydrogen atom replaces the electron with a muon of mass $m_\mu \approx 207\,m_e$ (same charge $-e$ , so the Coulomb attraction to the proton is unchanged in form). For the ground state ( $n=1$ ):

r_1^\mu = \frac{a_0}{207}, \qquad E_1^\mu = 207\times E_1^{(e)}

where $a_0 = 5.29\times10^{-11}\ \text{m}$ and $E_1^{(e)} = -13.6\ \text{eV}$ are the ordinary (electronic) hydrogen values.

Step 3 -- Evaluate the muonic Bohr radius. …

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Q.Obtain the first Bohr's radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon (μ−\mu^-μ−) of mass about 207me207m_e207me​ orbits around a proton].

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Q.Obtain the first Bohr's radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon ( $\mu^-$ ) of mass about $207m_e$ orbits around a proton].

Unlock everything free for 14 days