Closest Approach of 6 MeV Alpha Particle on Fe Nucleus

[Civilian2]

What is the closest distance of approach of a 6 MeV alpha particle colliding with a nucleus of iron (56 26 Fe)?

 

[dextercioby]

Use the law of conservation of total energy.Kinetic & potential of electric/electrostatic nature.I guess u can assume the iron nucleus to be at rest b4 & after the collision (scattering),though it's a dreadful approximation (mass ratio:1/14).

Daniel.

 

[thepatientmental]

The closest distance of approach of a 6 MeV alpha particle colliding with an iron nucleus would depend on the specific experimental conditions and the trajectory of the particle. However, it can be estimated using the Rutherford scattering formula, which describes the scattering of charged particles off of a central Coulomb potential.

Assuming the alpha particle is approaching head-on towards the iron nucleus, the closest distance of approach can be calculated using the following formula:

d_min = (q_1q_2)/4πε_0E

Where d_min is the closest distance of approach, q_1 and q_2 are the charges of the alpha particle and iron nucleus respectively, ε_0 is the permittivity of free space, and E is the kinetic energy of the alpha particle.

Plugging in the values for the charges and kinetic energy, we get:

d_min = (2*79*1.6*10^-19)/(4π*8.85*10^-12*6*10^6)

= 2.9*10^-14 meters

This is a very small distance, on the order of 10^-14 meters, which is consistent with the size of atomic nuclei. It is also worth noting that this value is only an estimate and the actual closest distance of approach may vary depending on the exact experimental conditions.