Question

Two protons in a star are involved in a head on collision. If the kinetic energy of each of these protons is 18 keV, what would be the distance of closest approach between them? (k=9*10⁹ Nm² C⁻²) [Ans: 4*10⁻¹⁴ m]

Answer 1

Dear Student,

◆ Answer -

d = 4×10^-14 m

◆ Explanation -

# Given -

E = 18 KeV

E = 18×10^3×1.6×10^-19 J

E = 2.88×10^-15 J

q1 = +1.6×10^-19 C

q2 = +1.6×10^-19 C

# Solution -

Closest distance of approach between two protons is calculated as -

d = k.q1.q2/E'

d = k.e^2/(2E)

d = 9×10^9 × (1.6×10^-19)^2 / (2 × 2.88×10^-15)

d = 4×10^-14 m

Therefore, closest distance of approach between two protons is 4×10^-14 m.

Thanks dear...