The time period of revolution of electron inits 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 :

Question

The wavelength of the light emitted by a hydrogen atom during a transition from $ n = 3 $ to $ n = 2 $ is $ 656.3 \, \text{nm} $. What is the energy of the photon emitted during this transition?

Answer 1

To solve the problem, we need to determine the frequency of revolution of the electron in its first excited state in a hydrogen atom. Given that the time period of revolution for the electron in its ground state is $1.6 \times 10^{-16}$ s, we'll use the following steps:

Step 1: Understanding the Relationship Between Time Period and Frequency

Frequency ($f$) is the reciprocal of the time period ($T$). Therefore, the frequency of revolution for the ground state is:

f = \frac{1}{T} = \frac{1}{1.6 \times 10^{-16}} \, \text{s}^{-1} = 6.25 \times 10^{15} \, \text{s}^{-1}

Step 2: Using Bohr's Model for Hydrogen Atom

According to Bohr's model, the relationship between the frequency of an electron in any orbit ($n$) and its time period can be derived based on the principal quantum number $n$. The frequency of revolution for any orbit is inversely proportional to the cube of the principal quantum number ($n^3$).

Thus, the frequency of the electron in the $n$-th orbit ($f_n$) is given by:

f_n = \frac{f_1}{n^3}

For the first excited state ($n=2$), we need to calculate:

f_2 = \frac{6.25 \times 10^{15}}{2^3} = \frac{6.25 \times 10^{15}}{8} = 7.8 \times 10^{14} \, \text{s}^{-1}

Conclusion: The frequency of revolution of the electron in its first excited state is $7.8 \times 10^{14} \, \text{s}^{-1}$.

Thus, the correct option is: 7.8 \times 10^{14}

Answer 2