Q.When an electric discharge is passed through hydrogen gas, the hydrogen molecules dissociate to produce excited hydrogen atoms. These excited atoms emit electromagnetic radiation of discrete frequencies which can be given by the general formula
ν̄ = 109677 [1/n_i^2 - 1/n_f^2]
What points of Bohr's model of an atom can be used to arrive at this formula? Based on these points derive the above formula giving description of each step and each term.

Bohr's model of the atom provides the theoretical framework to derive the given formula for the discrete frequencies of electromagnetic radiation emitted by hydrogen atoms.

The key points of Bohr's model used are:

1. Quantization of Orbits: Electrons revolve in certain stable, non-radiating orbits (stationary states) without emitting energy.
2. Quantization of Angular Momentum: The angular momentum of an electron in a stationary orbit is quantized, given by $L = mvr = n\frac{h}{2\pi}$, where $n$ is a positive integer (principal quantum number), $m$ is the electron's mass, $v$ is its speed, $r$ is the orbit radius, and $h$ is Planck's constant.
3. Energy Transitions: An atom emits or absorbs energy only when an electron makes a transition from one stationary orbit to another. The frequency ($\nu$) of the emitted or absorbed radiation is given by Bohr's frequency condition: $h\nu = E_{initial} - E_{final}$.

Derivation of the Formula:

1. Energy of an electron in the $n$-th orbit: For an electron in a stable circular orbit around a nucleus (with charge $+e$ for hydrogen), the electrostatic force provides the necessary centripetal force:

From Bohr's angular momentum quantization condition, $mvr = n\frac{h}{2\pi}$, which implies $v = \frac{nh}{2\pi mr}$. Substituting this into the force equation and solving for $r$ gives the radius of the $n$-th orbit:

The total energy $E_n$ of the electron in the $n$-th orbit is the sum of its kinetic energy ($KE = \frac{1}{2}mv^2$) and potential energy ($PE = -\frac{1}{4\pi\epsilon_0} \frac{e^2}{r}$). Using the force equation, $KE = \frac{e^2}{8\pi\epsilon_0 r}$. Thus,

Substituting the expression for $r_n$:

This formula gives the discrete energy levels of the hydrogen atom.

2. Bohr's Frequency Condition:

When an electron transitions from a higher energy initial state ($n_f$) to a lower energy final state ($n_i$), it emits a photon. The energy of this photon is $h\nu = E_{n_f} - E_{n_i}$.

$h\nu = \left(-\frac{m e^4}{8\epsilon_0^2 h^2 n_f^2}\right) - \left(-\frac{m e^4}{8\epsilon_0^2 h^2 n_i^2}\right)$ …

The hydrogen spectrum, characterized by discrete lines, was a significant challenge to classical physics. Rutherford's model, while explaining the atom's structure, failed to account for atomic stability and the observed discrete spectra. Niels Bohr addressed these issues by introducing a revolutionary model based on a few fundamental postulates. These postulates, when applied to the hydrogen atom, naturally lead to the quantization of energy levels and, consequently, to the formula for the discrete frequencies of emitted radiation.

The points of Bohr's model crucial for deriving the given formula are:

1. Stationary States: Electrons revolve around the nucleus in certain fixed, stable orbits without radiating energy. These orbits are called stationary states, and each state has a definite energy.
2. Quantization of Angular Momentum: Only those orbits are permitted for which the angular momentum of the electron is an integral multiple of $\frac{h}{2\pi}$ (where $h$ is Planck's constant). That is, $L = mvr = n\frac{h}{2\pi}$, where $n$ is a positive integer (1, 2, 3, ...), called the principal quantum number.
3. Energy Transitions: An atom radiates energy only when an electron makes a transition from a higher energy stationary state to a lower energy stationary state. The frequency ($\nu$) of the emitted radiation is given by the energy difference between the two states: $h\nu = E_{higher} - E_{lower}$.

Let's now derive the formula based on these points. We consider a hydrogen atom with a nucleus of charge $+e$ and an electron of charge $-e$ and mass $m$ revolving in a circular orbit of radius $r$.

1. Quantization of Angular Momentum According to Bohr's second postulate, the angular momentum $L$ of the electron is quantized:

where $n = 1, 2, 3, \dots$ is the principal quantum number.
From this, we can express the velocity $v$ of the electron:

2. Force Balance For the electron to remain in a stable circular orbit, the electrostatic attractive force between the electron and the nucleus must provide the necessary centripetal force. The electrostatic force ($F_e$) is given by Coulomb's law:

The centripetal force ($F_c$) required for circular motion is:

Equating these two forces:

3. Radius of Bohr Orbits Substitute the expression for $v$ from equation (1) into equation (2):

We can cancel $r^2$ from both sides and rearrange to solve for $r$:

This formula gives the radius of the $n$-th stationary orbit. For $n=1$, this is the Bohr radius, $a_0 = \frac{h^2 \epsilon_0}{\pi m e^2}$.

4. Total Energy of the Electron

The total energy $E_n$ of the electron in the $n$-th orbit is the sum of its kinetic energy ($KE$) and potential energy ($PE$).

* Kinetic Energy:

    From the force balance equation (2), we have $mv^2 = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r}$.
    So, $KE = \frac{1}{2} \left(\frac{1}{4\pi\epsilon_0} \frac{e^2}{r}\right) = \frac{e^2}{8\pi\epsilon_0 r}$.
*   **Potential Energy:**
    The potential energy of an electron at a distance $r$ from a proton is given by:

    The negative sign indicates that the electron is bound to the nucleus.
*   **Total Energy:**

    Now, substitute the expression for $r_n$ into the total energy formula:

This formula shows that the energy levels of the hydrogen atom are quantized, depending on the integer $n$. The negative sign indicates that the electron is bound to the nucleus; energy must be supplied to remove it.

5. Energy Difference and Frequency of Emitted Radiation

According to Bohr's third postulate, when an electron transitions from a higher energy state ($n_2$) to a lower energy state ($n_1$), it emits a photon with energy $h\nu$.

Let $E_2$ be the energy of the higher state ($n_2$) and $E_1$ be the energy of the lower state ($n_1$), where $n_2 > n_1$.

To find the frequency $\nu$:

6. Derivation of the Wave Number Formula The wave number $\bar{\nu}$ is defined as the reciprocal of the wavelength ($\lambda$) and is related to frequency by $\bar{\nu} = \frac{\nu}{c}$, where $c$ is the speed of light. …