The speed of electron revolving in the fourth orbit of a hydrogen-like atom or ion is 1094 km/s. The — Atomic Structure Chemistry Question

💡 Solution & Explanation

Speed $v_n = \frac{2.188 \times 10^6 Z}{n}$ m/s. Given $v_4 = 1094 \times 10^3$ m/s, $n=4$. $1.094 \times 10^6 = \frac{2.188 \times 10^6 \times Z}{4} \implies 4 \times 1.094 = 2.188 \times Z \implies 4.376 = 2.188 Z \implies Z = 2$. $Z=2$ corresponds to He$^+$. Therefore, correct answer is B.