Question: 17: Let T_{1} and T_{2} be the energy of an electron in the first and second excited states of hydrogen atoms, respectively. According to the Bohr’s model of an atom, the ratio T_{1}: T_{2} is
(1) 4: 9
(2) 9: 4
(3) 1: 4
(4) 4: 1
Answer: Option (2)
Explanation:
According to Bohr’s model, the total energy of an electron in the n^{\text{th}} orbit of a hydrogen atom is given by
T_n = -\frac{13.6}{n^{2}} \ \text{eV}.
The first excited state corresponds to n = 2,
and the second excited state corresponds to n = 3.
Thus, the energy of the electron in the first excited state is
T_1 = -\frac{13.6}{2^{2}} = -\frac{13.6}{4}.
The energy of the electron in the second excited state is
T_2 = -\frac{13.6}{3^{2}} = -\frac{13.6}{9}.
The ratio of the energies is
T_1 : T_2 = \frac{13.6/4}{13.6/9}.
T_1 : T_2 = \frac{9}{4}.
Hence, the ratio T_{1}: T_{2} is 9 : 4.