An -particle having kinetic energy 4.0 MeV is projected towards tin nucleus (Z = 50). Select the cor — Atomic Structure Chemistry Question

💡 Solution & Explanation

Distance of closest approach $r_0 = \frac{k(Ze)(2e)}{K} = \frac{9\times10^9 \times 50 \times 2 \times (1.6\times10^{-19})^2}{4 \times 10^6 \times 1.6 \times 10^{-19}} = 3.6 \times 10^{-14}$ m. P.E. at $9.0 \times 10^{-14}$ m is $\frac{k(100e^2)}{r} = 1.6$ MeV. Total energy = 4.0 MeV, so at $4.5 \times 10^{-14}$ m, P.E. = 3.2 MeV, giving K.E. = $4.0 - 3.2 = 0.8$ MeV. It cannot reach $2.0 \times 10^{-16}$ m since it turns around at $r_0$. Therefore, correct answer is A, B, C.