Physics · Atoms · NEET
At the closest point the alpha particle momentarily stops (velocity = 0) before reversing. So its kinetic energy there is zero. By conservation of energy, all the starting kinetic energy K has become electric potential energy between the two positive charges. That is why we write K = (1 / 4 pi e0) (2e)(Ze) / d and solve for d.
It depends only on the kinetic energy K, not on m and v separately. Write K = (1/2) m v^2, so d = 4Ze^2 / (4 pi e0 m v^2). For a fixed v, d is proportional to 1/m. For a fixed K, changing m alone does nothing because d = 2Ze^2 / (4 pi e0 K). Read the question carefully: NEET fixes v, so the answer is 1/m.
Impact parameter b is the sideways (perpendicular) offset of the incoming path from the nucleus line. Distance of closest approach d is how near the particle actually gets. Only for a head-on hit (b = 0) does the particle stop and reverse, giving the smallest possible d from the pure energy formula. When b is not zero the particle curves away and never reaches that head-on d.
An alpha particle is a helium nucleus with charge +2e (two protons). The target nucleus has charge +Ze where Z is its atomic number (gold has Z = 79). The electric potential energy between them is (1 / 4 pi e0)(2e)(Ze)/d, so the product of charges gives the factor 2Ze^2.
No. The closest approach only gives an upper limit for the nuclear size. In NCERT's 7.7 MeV example d is about 30 fm, but the real gold nucleus radius is about 6 fm. The particle turns back before touching the nucleus, so d is larger than the actual radius. Higher energy alphas get closer and give a better size estimate.
When an alpha-particle of mass 'm' moving with velocity 'v' bombards a heavy nucleus of charge 'Ze', its distance of closest approach from the nucleus depends on m as:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
d = 2Ze^2 / (4 pi e0 K), where K is the kinetic energy of the alpha particle, Z is the atomic number of the nucleus, e is the electron charge and e0 is the permittivity of free space.
Using d = 2Ze^2 / (4 pi e0 K) with Z = 79 and K = 7.7 MeV = 1.23 x 10^-12 J, NCERT gets d about 3.0 x 10^-14 m, that is about 30 fm. This is an upper limit for the gold nucleus size.
Zero. The particle momentarily comes to rest and reverses direction, so its kinetic energy at that instant is zero and all energy is electric potential energy.
Closer. Since d is proportional to 1/K, a higher kinetic energy K makes the distance of closest approach smaller, so the particle penetrates nearer to the nucleus.
Yes. A head-on hit has impact parameter b = 0, which gives the smallest possible closest distance for a given energy. Any sideways offset makes the particle turn away sooner.
