JEE Main20248 Apr 2024Morning ShiftPhysicsAtomic PhysicsActual
In an alpha particle scattering experiment distance of closest approach for the particle is 4.5 10⁻¹⁴ ~m . If target nucleus has atomic number 80 , then maximum velocity of - particle is _______ 10^5 ~m / s approximately. ( 1 4 ₀ =9 10^9 . SI unit, mass of particle .=6.72 10⁻²⁷ ~kg )
Correct answer
0
Step-by-step solution
aligned v & = 4 KZe ^2 mr _ min & = 4 9 10^9 80 6.72 10⁻²⁷ 4.5 10⁻¹⁴ 1.6 10⁻¹⁹ & =9.759 10²⁵ 1.6 10⁻¹⁹ & =156 10^5 ~m / s aligned