Q.It is found experimentally that 13.6 eV energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius and the velocity of the electron in a hydrogen atom.
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🔒 Start your 14-day free trial to unlock the full solution →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=n2πh,n=1,2,3,…
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=mvr, where m is the electron mass, v its speed, and r the orbit radius. So the quantization condition becomes:
mvr=n2πh
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.
The quantity 2πh appears so often that it has its own symbol: ℏ (h-bar). So the condition is often written as L=nℏ.
What it predicts
Combining this quantization with Newton's law for circular motion (centripetal force = Coulomb attraction) gives:
- Radius of the nth orbit: rn=n2a0, where a0=0.529A˚ is the Bohr radius (the smallest orbit, n=1).
- Energy of the nth orbit: En=−n213.6eV
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=nℏ
- Write the force balance: rmv2=r2kZe2 (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 2πh — 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:
λ=ph=mvh
For an electron in a circular orbit of radius r, the circumference is 2πr. For the wave to be stable — not cancelling itself out — the circumference must contain an integer number of wavelengths:
2πr=nλ,n=1,2,3,…
Substitute λ=h/(mv):
2πr=n⋅mvh
Rearrange:
mvr=n⋅2πh
That's it. The left side mvr is the angular momentum L. So:
L=nℏ,where ℏ=2πh
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:
rmv2=4πϵ01r2e2
Combine this with mvr=nℏ and solve for r and E:
rn=me24πϵ0ℏ2⋅n2=a0n2
En=−8ϵ02h2me4⋅n21=−n213.6 eV …
The key idea is that the 13.6 eV binding energy equals the magnitude of the total energy E of the electron in the ground state of hydrogen. For a circular orbit, the total energy is half the potential energy: E=−2rke2.
Step 1: Convert the binding energy to joules.
13.6 eV=13.6×1.6×10−19=2.176×10−18 J
Since E=−13.6 eV, we have ∣E∣=2rke2.
Step 2: Solve for the orbital radius r.
r=2∣E∣ke2=2×2.176×10−18(9×109)(1.6×10−19)2
r=4.352×10−189×2.56×10−29=4.352×10−182.304×10−28=5.29×10−11 m
Step 3: Find the velocity using mv2/r=ke2/r2, so v=mrke2. …
The problem uses the experimentally measured ionization energy (13.6 eV) to deduce the electron's orbital radius and velocity in the Bohr model. The key is equating the ionization energy to the total energy of the electron in the ground state, then using Bohr's quantization condition. The orbital radius comes out to 5.29×10−11 m and the velocity to 2.19×106 m/s.
The 13.6 eV given is not just any number — it is the ionization energy, the minimum energy needed to completely remove the electron from the proton's influence. In the Bohr model, this equals the negative of the total energy of the electron in the n=1 orbit. So the electron's total energy in the ground state is −13.6 eV.
Why does this help? Because in the Bohr model, the total energy, orbital radius, and velocity are all linked by simple relations. If we know one, we can find the others. The trick is to work in SI units consistently — convert eV to joules, then use Coulomb's law and Newton's second law alongside Bohr's angular momentum quantization.
Let's go step by step.
1. Convert the given energy to joules
The ionization energy is 13.6 eV. Since 1 eV=1.602×10−19 J:
Eion=13.6×1.602×10−19=2.179×10−18 J
This is the magnitude of the total energy ∣E∣ of the electron in the ground state. In the Bohr model, the total energy is negative (bound state), so:
E=−2.179×10−18 J
A common mistake is to forget the negative sign. The total energy of a bound electron is negative; the ionization energy is the positive amount needed to bring it to zero energy. So E=−13.6 eV, not +13.6 eV.
2. Recall the Bohr model relations for hydrogen
For an electron in a circular orbit around a proton, two equations hold:
- Newton's second law (Coulomb force provides centripetal acceleration):
4πε01r2e2=rmv2
- Bohr's quantization of angular momentum (for the ground state, n=1):
mvr=2πh
Here m=9.109×10−31 kg (electron mass), e=1.602×10−19 C, ε0=8.854×10−12 F/m, and h=6.626×10−34 J⋅s.
The total energy of the electron in the n-th Bohr orbit is:
En=−8ε02h2me4⋅n21
For n=1, this gives E1=−13.6 eV — which is exactly our starting point.
3. Use the total energy to find the radius
The total energy is the sum of kinetic and potential energies:
E=21mv2−4πε01re2
From the centripetal equation, we have rmv2=4πε01r2e2, which gives:
mv2=4πε01re2
So the kinetic energy K=21mv2=21⋅4πε01re2.
The potential energy U=−4πε01re2.
Therefore:
E=K+U=21⋅4πε01re2−4πε01re2=−21⋅4πε01re2
This is a neat result: the total energy is exactly half the potential energy (and the negative of the kinetic energy). Rearranging for r:
r=−21⋅4πε01Ee2
Since E is negative, r comes out positive. Plug in the numbers:
r=−21⋅4π(8.854×10−12)1⋅−2.179×10−18(1.602×10−19)2
First compute 4πε01=8.988×109 N⋅m2/C2. Then:
r=21×8.988×109×2.179×10−182.566×10−38 …
Method: Bohr Model for Hydrogen Atom
This problem uses the Bohr model of the hydrogen atom, which combines classical circular motion with quantised angular momentum. The key idea is that the electron orbits the proton in a circle, held by electrostatic attraction, but only certain orbits are allowed — those where the angular momentum is an integer multiple of 2πh.
Step 1: Write the energy condition
The total energy of the electron in the n=1 orbit (ground state) is given as 13.6 eV. This is the ionisation energy — the energy needed to remove the electron completely. In the Bohr model, the total energy is:
E=−2rke2
where k=4πϵ01=9×109 N m2/C2, e=1.6×10−19 C, and r is the orbital radius. The negative sign means the electron is bound.
Since ∣E∣=13.6 eV, we have:
2rke2=13.6 eV
Convert 13.6 eV to joules: 13.6×1.6×10−19=2.176×10−18 J.
Step 2: Solve for the orbital radius r
From the energy equation:
r=2Eke2
Substitute values:
r=2×2.176×10−18(9×109)(1.6×10−19)2
r=4.352×10−189×109×2.56×10−38
r=4.352×10−182.304×10−28
r=5.29×10−11 m
This is the Bohr radius (a0), a fundamental constant in atomic physics.
Step 3: Find the velocity v
The centripetal force is provided by the electrostatic attraction:
rmv2=r2ke2
Cancel one r:
mv2=rke2
So: …
Common Mistakes in Rutherford/Bohr Radius & Velocity Problems
This question is a classic — it tests whether you truly understand the link between energy, radius, and velocity in the Bohr model. Most students jump straight to formulas without connecting the given energy to the right physical quantity.
Mistake 1: Confusing the given 13.6 eV with kinetic energy
The most frequent error: students see 13.6 eV and immediately plug it into Ek=21mv2 to find velocity. But 13.6 eV is the total energy (or the ionization energy), not the kinetic energy.
Why this happens: The problem says "energy required to separate" — that's the binding energy, which equals the magnitude of the total energy of the electron in the ground state. In the Bohr model:
Etotal=−n213.6 eV
For n=1, Etotal=−13.6 eV. The kinetic energy is +13.6 eV, and the potential energy is −27.2 eV.
Never equate the given ionization energy directly to 21mv2. The kinetic energy is positive and equal in magnitude to the total energy, but the total energy itself is negative.
How to avoid: Always write down the three energy relations side by side before substituting numbers:
- Etotal=−2rke2 (or −13.6/n2 eV)
- Ek=2rke2=+13.6 eV (for n=1)
- Ep=−rke2=−27.2 eV
Mistake 2: Using the wrong value of e (electronic charge)
Students often use e=1.6×10−19 C but forget to square it when it appears as e2 in formulas. Or they mix up e with the charge on the nucleus (Ze), but here Z=1.
How to avoid: Write the formula with e2 explicitly, then substitute e=1.6×10−19 C and square it immediately. Keep units consistent — use SI units throughout.
Mistake 3: Forgetting to convert eV to joules
The given energy is in electronvolts, but the Coulomb constant k=9×109 and other quantities are in SI units. If you plug 13.6 directly without converting to joules, your radius will be off by a factor of 1019.
1 eV=1.6×10−19 J, so 13.6 eV=13.6×1.6×10−19=2.176×10−18 J.
How to avoid: Before starting any calculation, convert all energies to joules. Write the conversion step explicitly.
Mistake 4: Using the wrong formula for orbital radius
Some students try to derive radius from F=rmv2 alone, forgetting they need a second equation (either energy quantization or angular momentum quantization). Without Bohr's postulate, you cannot get a unique radius.
The correct approach: Use the total energy relation:
Etotal=−2rke2
Since ∣Etotal∣=13.6 eV=2.176×10−18 J, we have:
2rke2=2.176×10−18
Solve for r:
r=2×2.176×10−18ke2
r=2∣Etotal∣ke2
Mistake 5: Computing velocity from the wrong energy
After finding r, students sometimes use Ek=21mv2 but plug in the total energy value (negative) instead of the kinetic energy magnitude. …
Showing the 12 most recent of 29 on this concept.
- CBSE 2026Set 55/2/11 markMCQQ.In Bohr model of hydrogen atom, for large values of n, the distance between the consecutive orbits is proportional to (A) n (B) n (C) n2 (D) n3
›Reveal solutionSolution
In the Bohr model, the orbital radius scales as rn∝n2. For large n, the gap between consecutive orbits Δr=rn+1−rn behaves like 2n+1, which is proportional to n — so the answer is (B).
The Bohr model gives a beautifully simple picture of the hydrogen atom: electrons orbit the nucleus in fixed circular paths, with quantised angular momentum. The radius of the n-th orbit is
rn=πme2n2h2ε0or, more compactly,rn=a0n2,
where a0≈0.529A˚ is the Bohr radius. So the radius grows as n2.
The question asks about the distance between consecutive orbits — that is, the difference rn+1−rn — for large n. Many students instinctively think this difference is constant, or that it grows like n2 because the radii themselves do. But a difference between two quadratic terms behaves differently from either term alone. Let’s work it out.
- Write the radii for two neighbouring orbits:
rn=a0n2,rn+1=a0(n+1)2.
- The gap between them is
Δr=rn+1−rn=a0[(n+1)2−n2].
- Expand (n+1)2=n2+2n+1. Then
Δr=a0(n2+2n+1−n2)=a0(2n+1).
- For large n, the constant 1 becomes negligible compared to 2n. So
Δr≈2a0n.
Thus the spacing between consecutive orbits is proportional to n itself — not n2, not n, not n3. …
- CBSE 2026Set A1 markMCQQ.The radius of the lowest Bohr's orbit in hydrogen atom is r0. The radius of Bohr's second orbit is (A) r0 (B) 2r0 (C) 4r0 (D) r0/2
›Reveal solutionSolution
Bohr radii scale as n², so r₂ = 4 r₀.
In Bohr's model of hydrogen the radius of the n-th orbit is:
rn=n2r0
…
- CBSE 2026Set ANNUAL1 markMCQQ.If the first Bohr radius of hydrogen atom be R, then the radius of the third orbit is(a) 9R(b) R/3(c) 3R(d) R/9
›Reveal solutionSolution
Bohr radius of the nth orbit scales as n2, so the third orbit's radius is 9R.
In the Bohr model, the radius of the nth orbit of the hydrogen atom is
rn=n2r1
…
- CBSE 2026Set ANNUAL1 markMCQQ.Case study: Bohr's model addressed the instability of the Rutherford model by introducing quantization. The model is based on three postulates, which successfully explained the discrete line spectrum of hydrogen. According to the model, the radius of the nth stationary orbit is r_n ∝ n², and the total energy is E_n = −13.6 eV / n², when an electron jumps from a higher energy level (E_i) to a lower one (E_f), a photon of energy hν = E_i − E_f is emitted. Transitions ending at the n = 1 level form the Lyman series. According to Bohr's second postulate, which quantity is quantized?(a) Energy of electron(b) Orbital angular momentum(c) Linear momentum(d) Frequency of revolution
›Reveal solutionSolution
Bohr's second postulate quantizes the electron's orbital angular momentum in integer multiples of h/2π.
Bohr's second postulate states that an electron can revolve only in those orbits for which its orbital angular momentum is an integral multiple of ℏ=h/2π: …
- CBSE 2025Set ANNUAL1 markMCQQ.According to Bohr's hypothesis the following physical quantity is quantised(a) angular momentum(b) angular velocity(c) potential energy(d) momentum
›Reveal solutionSolution
Bohr's key postulate (beyond classical mechanics) was that only orbits where the electron's angular momentum is an integer multiple of h/2π are allowed.
Bohr's second postulate states that the electron can revolve only in those orbits for which its orbital angular momentum is an integral multiple of h/2π:
L=mvr=2πnh,n=1,2,3,…
…
- CBSE 2025Set ANNUAL1 markMCQQ.The radius of Bohr's stable orbit for hydrogen is r. The radius of Bohr's second orbit is(a) r(b) r/2(c) 2r(d) 4r
›Reveal solutionSolution
In the Bohr model, the radius of the nth orbit of hydrogen scales as n^2, so if the (first, ground-state) orbit has radius r, the second orbit has radius 4r.
Bohr's model gives the radius of the nth stationary orbit of hydrogen as:
r_n = n^2 r_1
where r_1 is the radius of the first (n=1) orbit. Taking the given "stable orbit" radius r to be r_1, the second orbit (n=2) has radius:
…
- CBSE 2025Set ANNUAL1 markQ.If radius of first electron orbit of hydrogen is r0, radius of second electron orbit of hydrogen is ______.
›Reveal solutionSolution
In the Bohr model, the radius of the nth orbit of hydrogen scales as rn=n2r0, where r0 is the first-orbit (Bohr) radius.
Bohr's model gives the radius of the nth stationary orbit of the hydrogen atom as
rn=n2r0 …
- CBSE 2024Set ANNUAL1 markQ.If the radius of first orbit of hydrogen atom is 0.5 x 10^-10 m, then the radius of its second orbit will be __________ m.
›Reveal solutionSolution
Bohr's model gives orbit radius proportional to n², so the second orbit's radius is 2² = 4 times the first orbit's radius.
In Bohr's model of the hydrogen atom, the radius of the nth orbit is:
rn=n2r1
…
- CBSE 2024Set ANNUAL1 markQ.What is Bohr's quantisation condition for the angular momentum of an electron in the second orbit ?
›Reveal solutionSolution
Bohr's second postulate: angular momentum is quantised as L=2πnh; for the second orbit, n=2.
Bohr's quantisation condition states that an electron can revolve only in those orbits for which its orbital angular momentum is an integral multiple of h/2π: …
- CBSE 2023Set 55/1/11 markMCQQ.The radius of the nth orbit in Bohr model of hydrogen atom is proportional to :(a) n21(b) n1(c) n2(d) n
›Reveal solutionSolution
In the Bohr model, the electron's orbit radius grows with the principal quantum number because higher orbits require more angular momentum and lower electrostatic attraction. The radius is proportional to n2.
The Bohr model treats the hydrogen atom as a miniature solar system where the electron orbits the nucleus in circular paths. But unlike planets, the electron can only occupy certain allowed orbits, determined by quantum conditions. The question asks how the orbital radius scales with the quantum number n.
The key insight is that two forces govern the electron's motion: the electrostatic attraction pulling it inward and the requirement that its angular momentum be quantized. Let me show you how these constraints lead to the radius formula.
The physics behind the orbit
For a stable circular orbit, the centripetal force must equal the electrostatic force:
rmv2=r2ke2
where m is the electron mass, v its speed, k is Coulomb's constant, and e the electron charge. This gives us one equation relating r and v.
Bohr's quantum condition provides the second equation. He postulated that angular momentum is quantized:
mvr=nℏ
where ℏ=2πh and n=1,2,3,… is the principal quantum number.
Deriving the radius dependence
- From the angular momentum condition, solve for v:
v=mrnℏ
- Substitute this into the force balance equation:
rm(mrnℏ)2=r2ke2
- Simplify the left side:
m2r3m⋅n2ℏ2=mr3n2ℏ2=r2ke2
- Multiply both sides by r3:
mn2ℏ2=ke2r
- Solve for r: r=mke2n2ℏ2 …
- CBSE 2023Set ANNUAL1 markQ.Write the mathematical form of Bohr's postulate regarding angular momentum of electron in atom. (Write the answer only)
›Reveal solutionSolution
Bohr's second postulate states that the angular momentum of the electron in a stationary orbit is an integral multiple of h/2π.
Bohr postulated that an electron can revolve only in those orbits for which its orbital angular momentum is quantized:
L=mvr=2πnh,n=1,2,3,…
…
- CBSE 2023Set ANNUAL1 markMCQQ.The Bohr model of atoms(1) assumes that the angular momentum of electrons is quantized(2) uses Einstein's photoelectric equation(3) predicts continuous emission spectra for atoms(4) predicts the same emission spectra for all types of atoms
›Reveal solutionSolution
Bohr's central postulate was that electrons can only occupy orbits where angular momentum is an integer multiple of h/2π.
Bohr postulated that an electron revolves only in those orbits for which its angular momentum is quantized: L=mvr=2πnh, n=1,2,3,…. This quantization condition (not option b, which is unrelated to Bohr's model; and not options c/d, since Bohr's model correctly predicts DISCRETE, …
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