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

According to Bohr's model, the highest kinetic energy is associated with the electron in the
  1. first orbit of H atom
  2. first orbit of He+
  3. second orbit of He+
  4. 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) $\mathrm{KE}_{2, \mathrm{Li}}{ }^{\mathrm{i}^{+}}=+13.6 \times \frac{3^2}{2^2}=13.6 \times \frac{9}{4} \mathrm{eV}$ !

Asked in: JEE Advanced 2024 (Paper 2)

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