An alpha particle of energy 5 MeV is moving forward for a head-on collision. The distance of closest approach from the nucleus of atomic number Z=50 is ×10⁻¹⁴ m. (k=9×10⁹ SI, e=1.6×10⁻¹⁹ C, 1 eV=1.6×10⁻¹⁹ J)

Question

Accepted Answer

(2) 2.88. Given: E=5 MeV=5×10⁶×1.6×10⁻¹⁹ J, Z=50, and the α-particle carries charge 2e. Idea: at the distance of closest approach all the kinetic energy has been converted into electrostatic potential energy. E=(k(2e)(Ze))/(r₀), so r₀=(k(2e)(Ze))/E. r₀=(9×10⁹×2×50×(1.6×10⁻¹⁹)²)/(5×10⁶×1.6×10⁻¹⁹)=(9×10⁹×100×1.6×10⁻¹⁹)/(5×10⁶). r₀=2.88×10⁻¹⁴ m, so the blank is 2.88.

An alpha particle of energy 5 MeV is moving forward for a head-on collision. The distance of closest approach from the nucleus of atomic number Z=50 is _____ ×10⁻¹⁴ m. (k=9×10⁹ SI, e=1.6×10⁻¹⁹ C, 1 eV=1.6×10⁻¹⁹ J)

Step-by-step solution

Why the other options are wrong