Arrange the kinetic energy of the electron in the following orbits in a hydrogen atom in increasing order.
(A) n = 4
(B) n = 2
(C) n = 1
(D) n = 3
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (D), (B), (C)
In a Bohr orbit the kinetic energy of the electron equals the magnitude of its total energy, so it is largest in the innermost orbit.
\( \displaystyle K_n = -E_n = \frac{13.6}{n^{2}}\,\mathrm{eV} \)
\( \text{(A) } n = 4: K = 0.85\,\mathrm{eV} \)
\( \text{(D) } n = 3: K = 1.51\,\mathrm{eV} \)
\( \text{(B) } n = 2: K = 3.4\,\mathrm{eV} \)
\( \text{(C) } n = 1: K = 13.6\,\mathrm{eV} \)
So the increasing order is (A), (D), (B), (C).