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]

Two protons in a star are involved in a head on collision. If the kine

1.	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=910⁹ Nm² C⁻²) [Ans: 410⁻¹⁴ m]
Answer» udent,◆ Answer -d = 4×10^-14 m◆ EXPLANATION -# Given -E = 18 KeV E = 18×10^3×1.6×10^-19 JE = 2.88×10^-15 J q1 = +1.6×10^-19 Cq2 = +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 mTherefore, closest distance of approach between two protons is 4×10^-14 m.Thanks DEAR...

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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]

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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=910⁹ Nm² C⁻²) [Ans: 410⁻¹⁴ m]