Q1 / 51 · JEE Advanced 2024 (Paper 2) · Chemistry

According to Bohr's model, the highest kinetic energy is associated with the electron in the

  • A
    first orbit of H atom
  • Correct answer
    B
    first orbit of He+
  • C
    second orbit of He+
  • D
    second orbit of Li2+

Solution

$K E=+13.6 \times \frac{Z^2}{n^2}$ (A) $\mathrm{KE}_{1,11}=+13.6 \times \frac{1^2}{1^2}=13.6 \mathrm{eV}$ (B) $\mathrm{KE}_{1, \mathrm{Hc}}=+13.6 \times \frac{2^2}{1^2}=13.6 \times 4 \mathrm{eV}$ (C) $\mathrm{KE}_{2, \mathrm{He}^{+}}=+13.6 \times \frac{2^2}{2^2}=13.6 \mathrm{eV}$ (D) $\mathrm{KE}_{2, \mathrm{Li}}{ }^{\mathrm{i}^{+}}=+13.6 \times \frac{3^2}{2^2}=13.6 \times \frac{9}{4} \mathrm{eV}$

Hence, the correct answer is option: 'B'.

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