Question

The model notes that electrons in atoms travel around a central nucleus in circular orbits and can only orbit stably at a distinct set of distances from the nucleus in such fixed circular orbits. These orbits are related to definite energies and are often referred to as energy shells or levels of energy.

Answer 1

The frequency is given by the formula:
$n = \dfrac{c}{\lambda }$ …… (1)
Where,
$n$ indicates frequency.
$c$ indicates speed of light.
$\lambda$ indicates wavelength of the photon.

Complete step by step answer:
A) From reference of Postulates of Bohr's
1. An atom has a number of stable orbits in which, without the emission of radiant energy, an electron will reside. Each orbit corresponds to a given level of energy.
2. An electron can spontaneously jump from one orbit (energy level ${E_1}$) to another orbit (energy level ${E_2}$)( ${E_2} > {E_1}$ ); then Planck's equation $\Delta E = {E_2} - {E_1} = hv$ gives the energy change $\Delta E$ in the electron jump.
3. In a circular orbit, an electron's motion is constrained in such a way that its angular momentum is an integral multiple of $\dfrac{h}{{2\pi }}$.
The shell was called a special surface around the nucleus that contained orbits of equal energy and radius. These shells are numbered as $1$, $2$, $3$, $4$ etc. from inside to outside and referred to as $K$, $L$, $M$, $N$ respectively.

B) A hydrogen atom initially in the ground state absorbs a photon which excites it to the $n = 4$ level.
Then the frequency of the photon will be,
\dfrac{1}{\lambda } = {R_{\text{H}}}\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{4^2}}}} \right] \\\ \Rightarrow \dfrac{1}{\lambda } = {R_{\text{H}}} \times \dfrac{{15}}{{16}} \\\ \Rightarrow \lambda = \dfrac{{16}}{{15{R_{\text{H}}}}} \\\

Frequency can be written as $\dfrac{c}{\lambda }$.
n = \dfrac{c}{\lambda } \\\ \Rightarrow n = \dfrac{{3 \times {{10}^8}}}{{\dfrac{{16}}{{15{R_{\text{H}}}}}}} \\\ \Rightarrow n = \dfrac{{3 \times {{10}^8}}}{{\dfrac{{16}}{{15 \times 1.1 \times {{10}^7}}}}} \\\ \Rightarrow n = 3.1 \times {10^{15}}\,{\text{Hz}} \\\
So, $f = 3.1 \times {10^{15}}\,{\text{Hz}}$.
Hence, the frequency of the photon is around $3.1 \times {10^{15}}\,{\text{Hz}}$.

Additional Information:
Bohr’s model: The hydrogen atom model of Bohr is based on three postulates: (1) an electron travels in a circular orbit around the nucleus, (2) the angular momentum of an electron is quantified in the orbit, and (3) the change in the energy of an electron when it makes a quantum leap from one orbit to another is often followed by a photon's emission or absorption. The Bohr model is semi-classical since it incorporates the electron orbit classical principle (postulate 1) with the new quantization concept (postulates 2 and 3).

Note:
- The Bohr model is significant because the quantization of electron orbits in atoms was the first model to postulate it.
- It therefore represents an early quantum theory that gave the development of modern quantum theory a start.
- To describe atomic states, it introduced the idea of a quantum number.