In a scattering experiment, a beam of  e− with an energy of 420 MeV scatters off an atomic nucleus. If the first minimum of the differential cross section is observed at a scattering angle of 45°, the radius of the nucleus (in fermi) is closest to

Correct option is C

Given:

• Energy of the electron beam: 420 MeV.
• Scattering angle of the first minimum: 45°.
• We are tasked to calculate the radius of the nucleus (R) in fermi.

Solution:

1. Relation between energy and wavelength:
   The energy (E) of the electron is related to its wavelength (λ) through the formula:
   λ = hc / E,
   where hc = 1240 MeV·fm.

   Substituting E = 420 MeV:
   λ = 1240 / 420 ≈ 3 fm.

2. Diffraction condition for the first minimum:
   The diffraction condition is given by:
   d × sin(θ) = n × λ,
   where:

   • d is the effective diameter of the nucleus,
   • n = 1 (for the first minimum),
   • θ = 45°.

   Rearranging to find d:
   d = λ / sin(θ).

3. Calculate the effective diameter (d):
   Substituting λ = 3 fm and sin(45°) = 1/√2 ≈ 0.707:
   d = 3 / 0.707 ≈ 4.24 fm.

4. Radius of the nucleus:
   The radius (R) is half the diameter (d):
   R = d / 2 = 4.24 / 2 ≈ 2.5 fm.

Conclusion:
The radius of the nucleus is (c) 2.5 fermi.