An α-particle of energy 5 MeV is scattered through 180⁰ by a fixed uranium nucleus.

An α-particle of energy 5 MeV is scattered through 180⁰ by a fixed uranium nucleus. The distance of the closest approach is of the order of

Options

(a) 1Å
(b) 10⁻¹⁰ cm
(c) 10⁻¹² cm
(d) 10⁻¹⁵ cm

Correct Answer:

10⁻¹² cm

Explanation:

No explanation available. Be the first to write the explanation for this question by commenting below.

admin:

Related Questions

  1. Complete the reaction: ₀n¹+ ₉₂U²³⁵→₅₆Ba¹⁴⁴+……..+3n
  2. A proton carrying 1 MeV kinetic energy is moving in a circular path of radius
  3. In which of the following systems will the radius of the first orbit (n=1) be minimum?
  4. A particle of mass 1 mg has the same wavelength as an electron moving with a velocity
  5. A piece of marble is projected from the earth’s surface with velocity of 50 m/s