An -particle of momentum kg ms is projected towards the nucleus of an atom of an element. If the dis — Atomic Structure Chemistry Question
Question
An $\alpha$-particle of momentum $3.2 \times 10^{-20}$ kg ms$^{-1}$ is projected towards the nucleus of an atom of an element. If the distance of closest approach of $\alpha$-particle is $1.5 \times 10^{-13}$ m, the atomic number of element is (Mass of $\alpha$-particle = 4 amu, charge on electron = $1.6 \times 10^{-19}$ coulomb, $N_A = 6 \times 10^{23}$)
💡 Solution & Explanation
Mass of $\alpha$-particle $m = \frac{4}{6 \times 10^{23}} \text{ g} = \frac{2}{3} \times 10^{-26}$ kg. Kinetic Energy $K = \frac{p^2}{2m} = \frac{(3.2 \times 10^{-20})^2}{2 \times (2/3) \times 10^{-26}} = \frac{10.24 \times 10^{-40}}{1.33 \times 10^{-26}} = 7.68 \times 10^{-14}$ J. At the distance of closest approach, K.E. = P.E. = $\frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r}$. $7.68 \times 10^{-14} = 9 \times 10^9 \frac{2Z(1.6 \times 10^{-19})^2}{1.5 \times 10^{-13}} \implies 7.68 \times 10^{-14} = 30.72 \times 10^{-16} Z \implies Z = 25$. Therefore, correct answer is 0025.