Answer:
a) r_min = 5.69*10^-14
b) Fe_max = 11.26 N
Step-by-step explanation:
Given:-
- The kinetic energy of the alpha particle, K.E = 4.00 MeV
- The number of protons of Gold nucleus, n = 79 protons
Find:-
a) Determine the distance of closest approach of the alpha particle to the gold nucleus
b) Determine the maximum force exerted on the alpha particle. magnitude
Solution:-
- There are no fictitious forces unbalanced forces acting on the system consisting of an alpha particle and Gold nucleus. Then we can say that the total energy of the system is conserved.
- The alpha particle with the initial kinetic energy of 4.0 MeV approaches the Gold nucleus and with decreasing separation the electrostatic forces (Fe) becomes pre-dominant due to + charges repel.
- All the kinetic energy of the alpha particle is converted into the Electrostatic potential energy ( Ue ) when it reaches a distance of closest approach r_min.
- Apply conservation of energy:
K.E = Ue
K.E = k*(q_alpha)*(q_gold) / r_min
Where,
q_alpha: The net charge on the alpha particle = 2e
q_gold: The net charge on the gold nucleus = ne = 79e
k : Coulomb's constant = 8.99*10^9
r_min = k*(q_alpha)*(q_gold) / K.E
r_min = (8.99*10^9)*(2e)*(79e) / (4*10^6)e
r_min = 5.69*10^-14 m
- The maximum electrostatic force ( Fe ) acts on the approaching alpha particle when the the alpha particle is at its "distance of closest approach" r_min = 5.69*10^-14.
- The electrostatic force ( Fe ) is given as:
Fe = K.E / r_min
Fe = (4*10^6)*(1.602*10^-19) / ( 5.69*10^-14)
Fe = 11.26 N