Q.The deuteron is bound by nuclear forces just as H-atom is made up of p and e bound by electrostatic forces. If we consider the force between neutron and proton in deuteron as given in the form of a Coulomb potential but with an effective charge e′: F=4πε01re′2, estimate the value of (e′/e) given that the binding energy of a deuteron is 2.2 MeV.
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Rutherford Scattering Distance – From Intuition to Precision
Imagine you are firing a tiny, fast bullet at a large, heavy cannonball hidden inside a big cloud of cotton. Most bullets zip right through the cotton, barely slowing down. But a few bullets come very close to the cannonball itself. Those bullets get deflected sharply, sometimes even bouncing back.
The Rutherford scattering distance is the answer to this question: How close did that bullet get to the cannonball before it turned around?
In the real experiment, the "bullet" is an alpha particle (a helium nucleus, positively charged), the "cannonball" is the gold nucleus (also positively charged, and very heavy), and the "cotton" is the mostly empty space inside the gold atom. The alpha particle and the gold nucleus repel each other because both are positive. The closer the alpha particle gets, the stronger the repulsion.
The Intuitive Picture
Think of a ball rolling up a steep hill. The ball starts with some speed (kinetic energy). As it climbs, it slows down because gravity is pulling it back. At the very top of its climb, it stops for an instant — all its kinetic energy has been converted into gravitational potential energy. Then it rolls back down.
The alpha particle does the same thing, but with electric repulsion instead of gravity. It approaches the nucleus, slows down, stops at the closest possible point, and then flies back the way it came.
That closest point — the distance of closest approach — is the Rutherford scattering distance. It is the distance at which the alpha particle's initial kinetic energy is completely converted into electrostatic potential energy.
This distance is not the radius of the nucleus. It is the distance at which the alpha particle would just touch the nucleus if the nucleus were a point charge. In reality, the alpha particle never actually reaches the nucleus — it turns around before that.
The Precise Statement
Let an alpha particle with charge +2e and mass m approach a gold nucleus with charge +Ze (where Z=79 for gold). The alpha particle starts from very far away with initial kinetic energy K=21mv2.
At the distance of closest approach, call it r0, the alpha particle's speed becomes zero. All its kinetic energy has become electrostatic potential energy:
K=4πε01⋅r0(2e)(Ze)
Solving for r0:
r0=4πε01⋅K2Ze2
This is the Rutherford scattering distance (also called the distance of closest approach in a head-on collision).
What It Tells Us
- If the alpha particle hits the nucleus head-on, it comes exactly this close before reversing direction.
- If it misses slightly, it comes closer than r0? No — it comes less close. The head-on collision gives the minimum possible distance of closest approach for a given initial energy. Any sideways motion means the particle never gets as close.
- If the initial kinetic energy is larger, r0 becomes smaller — the alpha particle can punch closer to the nucleus before being stopped. …
Why this formula?
Rutherford Scattering: Why the Distance of Closest Approach Formula Works
The distance of closest approach — often denoted d0 or r0 — is the minimum separation between an alpha particle and the nucleus in a head-on collision. It's a beautiful example of energy conservation doing all the heavy lifting.
The Physical Picture
Imagine an alpha particle (charge +2e) fired straight at a gold nucleus (charge +Ze). As it approaches, the Coulomb repulsion slows it down. At the point of closest approach, the alpha particle's radial velocity becomes zero — it stops moving toward the nucleus, and is about to turn around and fly back.
At that instant, all the kinetic energy it had at infinity has been converted into electrostatic potential energy. No other forces are at play (gravity is negligible, and we're far from the nuclear force range).
The Derivation in One Step
Let the alpha particle have initial kinetic energy K=21mv2 at a large distance (where potential energy is zero). At the distance of closest approach r0, its speed is zero, so kinetic energy is zero. Energy conservation gives:
21mv2=4πϵ01⋅r0(2e)(Ze)
r0=4πϵ01⋅K2Ze2
That's it. The formula is a direct consequence of energy conservation in a pure Coulomb field.
Why This Makes Physical Sense
- Higher kinetic energy → the alpha particle can push closer before being stopped → r0 is smaller.
- Higher nuclear charge Z → stronger repulsion → the alpha stops farther away → r0 is larger.
- The factor 2Ze2 comes from the product of charges: (2e)(Ze)=2Ze2.
This is the head-on distance. For non-head-on collisions (nonzero impact parameter), the distance of closest approach is larger because some energy remains in the perpendicular component of motion. The general formula involves the impact parameter b and scattering angle θ, but the head-on case gives the absolute minimum possible approach.
A Common Misconception …
Treat the neutron-proton pair like a hydrogen-like (Bohr) bound state, but with the electron charge e replaced by an effective charge e′ and the electron mass replaced by the reduced mass of the two nucleons.
Bohr scaling: the ground-state binding energy of a hydrogen-like system is E∝μ(charge)4, where μ is the reduced mass.
Reduced mass of the deuteron: μ=mp+mnmpmn≈2mp=21836me=918me.
Scale from hydrogen (EH=13.6 eV, electron mass me, charge e): …
Modelling the deuteron as a hydrogen-like bound state with effective charge e′ and reduced mass μ=mp/2=918me gives ee′≈3.6.
The analogy
In the hydrogen atom an electron (mass me, charge e) is Coulomb-bound to the proton with ground-state binding energy EH=13.6 eV. The stem tells us to treat the neutron-proton force in the deuteron as a Coulomb-type interaction with an effective charge e′. So the deuteron is a hydrogen-like system with two changes: the charge e→e′, and the light electron mass is replaced by the reduced mass of the two heavy nucleons.
Step 1 — Bohr energy scaling
For any hydrogen-like bound state the ground-state energy is
E=21μc2α2,α=4πε0ℏc(charge)2,
so E∝μ(charge)4, where μ is the reduced mass of the bound pair.
Step 2 — Reduced mass of the deuteron
The neutron and proton have nearly equal masses ≈mp, so
μ=mp+mnmpmn≈2mp=21836me=918me.
Step 3 — Scale the binding energy
Replacing me→μ and e→e′, …
Method: Scaling a Bound-State Energy Between Two "Hydrogen-Like" Systems (Bohr-Model Scaling)
This method applies whenever a problem asks you to model an unfamiliar two-body bound system (here, the deuteron) as if it were a hydrogen-like atom with a different charge and/or different constituent masses, and to find an unknown parameter (like an effective charge) by matching binding energies.
Steps
Step 1: Recognise the two-body Coulomb-bound-state analogy
Any two particles held together purely by an inverse-square attractive force behave like a hydrogen-like atom. The Bohr-model ground-state binding energy for such a system depends only on the reduced mass of the pair and the fourth power of the interaction charge:
E∝μ(charge)4
Step 2: Compute the reduced mass of the actual bound system
For two particles of mass m1 and m2, the reduced mass is
μ=m1+m2m1m2
Use the real masses of the particles in the problem (here, the neutron and proton) — not the electron mass used in ordinary hydrogen.
Step 3: Set up the scaling ratio against the known hydrogen-atom result
Since you know hydrogen's own ground-state binding energy EH=13.6 eV (mass me, charge e), write the ratio of the new system's energy to hydrogen's:
EHEnew=(meμnew)(echargenew)4 …
- AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQQ.Find the impact parameter of a particle of energy 10 MeV while approaching the gold nucleus, if scattered with 60∘? (Charge of electron =1.6×10−19C, Atomic Number of gold =79) (A) 1.33 fermi (B) 11.33 fermi (C) 1133 fermi (D) 1333 fermi
›Reveal solutionSolution
This tests the Rutherford scattering impact-parameter formula, relating the closeness of approach (impact parameter b) of a charged projectile to its scattering angle. Answer: 11.33 fermi.
Concept and Intuition
In Rutherford's alpha-scattering picture, a projectile aimed with a smaller impact parameter b passes closer to the nucleus and experiences a stronger Coulomb repulsion, so it scatters through a larger angle θ. The exact relationship (derived from the hyperbolic Coulomb trajectory) is b=4πε0EkZe2cot(θ/2), where Ek is the kinetic energy of the incoming particle and Z is the nuclear charge number of the target (gold here).
Step-by-Step Solution
- Formula: b=4πε0EkZe2cot(θ/2)=EkkZe2cot(θ/2), with k=4πε01=9×109 N·m²/C².
- Data: Z=79, e=1.6×10−19 C so e2=2.56×10−38 C², Ek=10 MeV =10×1.6×10−13=1.6×10−12 J, θ=60∘⇒θ/2=30∘, cot30∘=3.
- kZe2=9×109×79×2.56×10−38=1.82×10−26. …
- AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQQ.In Rutherford α scattering experiment when a particle approaches with an impact parameter zero, its angle of scattering is (A) 0o (B) 2π (C) π (D) 32π
›Reveal solutionSolution
Zero impact parameter means a head-on collision course; Coulomb repulsion sends the alpha particle straight back the way it came, giving a scattering angle of π (180°).
Concept and Intuition
The impact parameter is the perpendicular distance between the incoming particle's original straight-line path and the nucleus. When it is zero, the particle is aimed directly at the nucleus. As it approaches, the repulsive Coulomb force decelerates it until it momentarily stops (at the distance of closest approach) and then is pushed directly backward along the same line — a complete reversal.
Step-by-Step Solution
- Impact parameter b=0 means the incoming trajectory points directly at the nucleus (no perpendicular offset).
- The Coulomb repulsion acts entirely along this line of approach, decelerating the particle to rest at closest approach. …
- AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQQ.An alpha particle of energy K MeV is moving towards a nucleus of atomic number Z. The distance of closest approach of the alpha particle to the nucleus in metres is (A) 7.2×10−16 KZ (B) 3.84×10−16 KZ (C) 14.4×10−16 KZ (D) 28.8×10−16 KZ
›Reveal solutionSolution
At closest approach all kinetic energy is converted to Coulomb potential energy; plugging in constants gives the numeric coefficient 28.8×10−16.
Concept and Intuition
As an alpha particle (charge +2e) approaches a nucleus (charge +Ze) head-on, it slows down due to Coulomb repulsion until, at the distance of closest approach d, all its kinetic energy has been converted into electrostatic potential energy. This is the classic Rutherford scattering geometry.
Step-by-Step Solution
- Energy conservation: E=4πε01d(2e)(Ze)=d2kZe2, where k=4πε01=9×109 N m2C−2.
- Rearranged: d=E2kZe2.
- Compute the constant: 2ke2=2×9×109×(1.6×10−19)2=4.608×10−28 J·m.
- Convert E from MeV to Joules: E=K×1.6×10−13 J. …
- AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQQ.The distance of closest approach of an alpha particle to a nucleus when the alpha particle moves towards the nucleus with linear momentum P is d. The distance of closest approach of alpha particle to nucleus, if the linear momentum of the alpha particle is 1.5 P (A) 32d (B) 23d (C) 94d (D) 49d
›Reveal solutionSolution
This tests how the distance of closest approach in Rutherford scattering depends on the momentum of the incoming particle. Answer: 94d.
Concept and Intuition
At the distance of closest approach, all of the alpha particle's initial kinetic energy has converted into electrostatic potential energy as it is repelled by the nucleus: KE=dkZe2, giving d=KEkZe2. Since kinetic energy in terms of momentum is KE=2mP2, we get d∝P21 — the closest approach distance shrinks quadratically as momentum increases, since a faster/more energetic particle penetrates further against the repulsive Coulomb force before turning back.
Step-by-Step Solution
- d=KEkZe2=P2kZe2⋅2m, so d∝P21.
- For momentum P: d=P2C for some constant C. …
- AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQQ.Assertion (A): The impact parameter for scattering of α-particles by 180° is zero. Reason (R): Zero impact parameter means that the α-particles tend to hit the center of the nucleus. (A) Both A and R are true and R is a correct explanation for A (B) Both A and R are true but R is not a correct explanation for A (C) A is true, R is false (D) A is false, R is true
›Reveal solutionSolution
In Rutherford scattering, a 180° deflection genuinely requires a head-on approach (zero impact parameter), and "zero impact parameter" by definition means the particle's trajectory (if undeflected) would pass through the nucleus's centre — so R does correctly explain A.
Concept and Intuition
The impact parameter b is the perpendicular distance between the initial straight-line path of an approaching alpha particle and the (extended) line through the nucleus's centre. Larger b means a more glancing collision (small deflection); b=0 means the particle is aimed directly at the nucleus, leading to a maximal deflection of 180° (it decelerates, stops, and is repelled straight back along its original path).
Step-by-Step Solution
- Assertion (A): "impact parameter for 180° scattering is zero" — this is a well-established result of Rutherford's scattering formula (b→0 as scattering angle θ→180°), so A is TRUE. …
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