Imagine removing one electron from $\text{He}^4$ and $\text{He}^3$. Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why.

Imagine removing one electron from He⁴ and He³. Their energy levels, as worked out on the basis of Bohr model

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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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Bohr energy levels depend on the nuclear charge $Z$ and the reduced mass $\mu$ . For $\text{He}^4$ and $\text{He}^3$ , $Z=2$ is identical and their reduced masses differ by only about $0.0045\%$ , so the levels are nearly the same.

Removing one electron from either $\text{He}^4$ or $\text{He}^3$ leaves a one-electron ion, $\text{He}^+$ , for which the Bohr model is exact:

E_n = -\frac{\mu\,Z^2 e^4}{8\epsilon_0^2 h^2}\cdot\frac{1}{n^2}.

1. The charge is the same. Both nuclei carry $Z = 2$ , so $Z^2$ is identical. The only quantity that can differ is the reduced mass

\mu = \frac{m_e M}{m_e + M},

which depends on the nuclear mass $M$ .

2. Reduced mass is very close to $m_e$ . Since $M \gg m_e$ ,

\mu \approx m_e\left(1 - \frac{m_e}{M}\right).

With $M_4 \approx 4.0026\,\text{u}$ , $M_3 \approx 3.0160\,\text{u}$ and $m_e \approx 5.486\times10^{-4}\,\text{u}$ :

\frac{m_e}{M_4} \approx 1.37\times10^{-4},\qquad \frac{m_e}{M_3} \approx 1.82\times10^{-4}.

3. The isotope difference. The fractional difference in $\mu$ between the two isotopes is the difference of these two small corrections, not their size: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KEAM 2026Set eng-2026-04194 marksMCQ
$0.53\ \overset{\circ}{A}$
›Reveal solutionSolution

Orbital radius grows as $n^2$ , so the 3rd orbit radius is $0.53\times9=4.77\,\text{\AA}$ .

In the Bohr model, the radius of the $n$ th orbit of hydrogen is
$r_n=r_1\,n^2,$
with $r_1=0.53\,\text{\AA}$ .

For $n=3$ : …
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KEAM 2026Set eng-2026-04204 marksMCQ
$v$
›Reveal solutionSolution

In the Bohr model $v_n=\dfrac{v_1}{n}$ , so the third-orbit speed is $v/3$ .

Reasoning. The speed of the electron in the $n$ -th orbit is $v_n\propto\dfrac{Z}{n}$ . For hydrogen with first-orbit speed $v$ , …
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KEAM 2026Set eng-2026-04214 marksMCQ
$5^{th}$
›Reveal solutionSolution

Bohr angular momentum $L_n=nh/2\pi$ ; difference $=h/\pi$ .

Bohr's quantization gives $L_n = \dfrac{nh}{2\pi}$ . …
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KEAM 2026Set eng-2026-04224 marksMCQ
$v$
›Reveal solutionSolution

In Bohr's hydrogen atom the orbital radius scales as $n^2$ and the orbital velocity as $1/n$ . …

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KEAM 2025Set pha-2025-0429F4 marksMCQ
$2^{nd}$
›Reveal solutionSolution

Bohr's angular momentum quantization gives $L = nh/2\pi$ ; the second orbit ( $n=2$ ) has $L = h/\pi$ .

Bohr's postulate quantizes the electron's orbital angular momentum as

$L = \frac{nh}{2\pi}.$ …
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KEAM 2024Set eng-2024-06054 marksMCQ
$Q. The ratio of the velocities of the electron in the second, third and fourth Bohr's orbits of hydrogen atom is (A) 3 : 2 : 1 (B) 1 : 2 : 3 (C) 1 : 4 : 9 (D) 6 : 4 : 3 (E) 9 : 4 : 1$
›Reveal solutionSolution

$v_n \propto 1/n \Rightarrow$ ratio $6:4:3$ for the 2nd, 3rd, 4th orbits.

The speed of an electron in the $n$ -th Bohr orbit is
$v_n = \frac{e^2}{2\varepsilon_0 h n} \propto \frac{1}{n}.$
For $n = 2,3,4$ : …
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KEAM 2024Set eng-2024-06074 marksMCQ
$\infty$
›Reveal solutionSolution

Number of de Broglie waves in the $n$ th orbit $=n$ ; third orbit $\Rightarrow 3$ .

Bohr's quantisation condition $mvr=\dfrac{nh}{2\pi}$ is equivalent to $2\pi r=n\lambda$ (de Broglie relation). This means exactly $n$ complete electron waves fit into the $n$ th or …
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KEAM 2022Set eng-2022-P1-A14 marksMCQ
$-3.4$
›Reveal solutionSolution
The energy −3.4 eV corresponds to n = 2, giving angular momentum L = 2h/2π = h/π.

Concept and Intuition

In the Bohr model, hydrogen's total energy is E_n = −13.6/n² eV, and the quantized angular momentum is L = nh/2π. Finding n from the energy fixes the angular momentum.

Step-by-Step Solution

Set E_n = −13.6/n² = −3.4 eV.

Solve: n² = 13.6/3.4 = 4, so n = 2. …
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Q.Imagine removing one electron from $\text{He}^4$ and $\text{He}^3$ . Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why.

Concept understanding — Bohr Model Quantization

Why does an electron not spiral into the nucleus?

The radical idea: allowed orbits only

What this means physically

What it predicts

The key insight for exams

Why this formula?

Why Angular Momentum is Quantised in the Bohr Model

The core problem Bohr faced

The de Broglie wavelength argument (the cleanest derivation)

Why this fixes the energy levels

Method: The General Isotope-Shift Formula (Reduced-Mass Approximation)

Steps

Concept and Intuition

Step-by-Step Solution

Q.Imagine removing one electron from He4\text{He}^4He4 and He3\text{He}^3He3. Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why.

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Q.Imagine removing one electron from $\text{He}^4$ and $\text{He}^3$ . Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why.

Unlock everything free for 14 days