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WBJEEAccording to Bohr's model, the highest kinetic energy is associated with the electron...
According to Bohr's model, the highest kinetic energy is associated with the electron in the
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) \(\mathrm{KE}_{2, \mathrm{Li}}{ }^{\mathrm{i}^{+}}=+13.6 \times \frac{3^2}{2^2}=13.6 \times \frac{9}{4} \mathrm{eV}\)
(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) \(\mathrm{KE}_{2, \mathrm{Li}}{ }^{\mathrm{i}^{+}}=+13.6 \times \frac{3^2}{2^2}=13.6 \times \frac{9}{4} \mathrm{eV}\)
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