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An alpha particle having kinetic energy 10Mev.an atom having atomic mass 35.calculate its distance of closest approach

Answer»

10.08 fermiExplanation:At it's closest approach,  where, q₁ = charge on ALPHA particle = 2 × 1.6 × 10⁻¹⁹               q₂ = charge on NUCLEUS = 35 × 1.6 × 10⁻¹⁹               r = DISTANCE of closest approach               K = 1/4π∈₀Hence,∴



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