JEE PYQ: Atomic Physics Question 36
Question 36 - 2020 (07 Jan Shift 1)
The time period of revolution of electron in its ground state orbit in a hydrogen atom is $1.6 \times 10^{-16}$ s. The frequency of revolution of the electron in its first excited state (in s$^{-1}$) is:
(1) $1.6 \times 10^{14}$
(2) $7.8 \times 10^{14}$
(3) $6.2 \times 10^{15}$
(4) $5.6 \times 10^{12}$
Show Answer
Answer: (2)
Solution
For first excited state $n = 2$. Time period $T \propto \frac{n^3}{Z^2}$. $\frac{T_2}{T_1} = \frac{n_2^3}{n_1^3} = 8$. $T_2 = 8T_1 = 8 \times 1.6 \times 10^{-16}$ s. Frequency $\nu = \frac{1}{T_2} = \frac{1}{8 \times 1.6 \times 10^{-16}} \approx 7.8 \times 10^{14}$ Hz.
BETA
