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Bohr Model of Hydrogen Atom

Bohr Model Of The Hydrogen Atom (Postulates)

To overcome the limitations of Rutherford's model (atomic stability and discrete spectra), Niels Bohr proposed a new model of the hydrogen atom in 1913, incorporating ideas from quantum theory. Bohr's model successfully explained the spectrum of hydrogen and provided a theoretical basis for the quantised energy levels of atoms.

Bohr’s Postulates

Bohr's model is based on the following postulates:

  1. Electron Orbits: Electrons in an atom revolve around the nucleus in certain definite circular paths called stationary orbits or allowed orbits. These orbits are stable, and the electron does not radiate energy while revolving in these orbits. This postulate contradicted classical electromagnetic theory, which predicted that an accelerating electron should radiate energy.
  2. Quantization of Angular Momentum: Only those orbits are allowed for which the angular momentum of the electron is an integral multiple of $\frac{h}{2\pi}$ or $\hbar$. If $m_e$ is the mass of the electron, $v_n$ is its speed in the $n$-th orbit, and $r_n$ is the radius of the $n$-th orbit, the angular momentum $L_n$ is given by:

    $ L_n = m_e v_n r_n = n \frac{h}{2\pi} = n\hbar $

    Where $n$ is a positive integer ($n=1, 2, 3, ...$), called the principal quantum number. Each value of $n$ corresponds to a specific allowed orbit. The state with $n=1$ is called the ground state, and states with $n>1$ are called excited states.
  3. Energy Transitions: An atom radiates energy only when an electron jumps from a higher energy stationary orbit to a lower energy stationary orbit. The energy is emitted in the form of a photon, and the energy of the emitted photon is equal to the difference in the energies of the two orbits. If an electron jumps from an initial orbit with energy $E_i$ (corresponding to quantum number $n_i$) to a final orbit with energy $E_f$ (corresponding to quantum number $n_f$, where $n_i > n_f$), the frequency ($\nu$) of the emitted photon is given by:

    $ h\nu = E_i - E_f $

    Conversely, an atom absorbs energy only when an electron jumps from a lower energy orbit to a higher energy orbit. The energy absorbed is also in the form of a photon of energy $h\nu = E_f - E_i$ (where $E_f > E_i$).

These postulates introduced the concept of quantised orbits and energy levels in atoms, explaining the discrete nature of atomic spectra.

Energy Levels ($ E_n = -13.6/n^2 $ eV)

Using Bohr's postulates, we can derive the allowed radii and energies of the electron in the hydrogen atom ($Z=1$).

From the first postulate and classical mechanics, the electrostatic force provides the centripetal force for circular motion in the $n$-th orbit:

$ \frac{1}{4\pi\epsilon_0} \frac{e^2}{r_n^2} = \frac{m_e v_n^2}{r_n} \quad \ldots (A) $

From the second postulate (quantization of angular momentum):

$ m_e v_n r_n = n\hbar \quad \ldots (B) $

We can solve these two equations for $v_n$ and $r_n$. From (B), $v_n = \frac{n\hbar}{m_e r_n}$. Substitute this into (A):

$ \frac{1}{4\pi\epsilon_0} \frac{e^2}{r_n^2} = \frac{m_e}{r_n} \left(\frac{n\hbar}{m_e r_n}\right)^2 = \frac{m_e}{r_n} \frac{n^2\hbar^2}{m_e^2 r_n^2} = \frac{n^2\hbar^2}{m_e r_n^3} $

Rearranging to solve for $r_n$:

$ \frac{e^2}{4\pi\epsilon_0} \frac{1}{r_n^2} = \frac{n^2\hbar^2}{m_e r_n^3} $

$ r_n^3 \frac{e^2}{4\pi\epsilon_0} = n^2\hbar^2 r_n^2 / m_e $

Assuming $r_n \neq 0$, we can divide by $r_n^2$:

$ r_n \frac{e^2}{4\pi\epsilon_0} = \frac{n^2\hbar^2}{m_e} $

$ r_n = \left(\frac{4\pi\epsilon_0 \hbar^2}{m_e e^2}\right) n^2 $

The term in the parenthesis is a constant. This gives the allowed radii of the electron orbits, which are proportional to the square of the principal quantum number ($r_n \propto n^2$). The radius of the first Bohr orbit ($n=1$) is called the Bohr radius, $a_0$.

$ a_0 = \frac{4\pi\epsilon_0 \hbar^2}{m_e e^2} $

Substituting the values of constants ($\epsilon_0 = 8.854 \times 10^{-12} \, C^2/(N \cdot m^2)$, $\hbar = 1.055 \times 10^{-34} \, J \cdot s$, $m_e = 9.109 \times 10^{-31} \, kg$, $e = 1.602 \times 10^{-19} \, C$):

$ a_0 \approx 0.529 \times 10^{-10} \, m = 0.0529 \, nm $

So, the radii of the allowed orbits are $r_n = a_0 n^2$.

Now, let's find the energy levels. The total energy of the electron in the $n$-th orbit is $E_n = -\frac{1}{8\pi\epsilon_0} \frac{e^2}{r_n}$ (for $Z=1$).

Substitute the expression for $r_n$:

$ E_n = -\frac{1}{8\pi\epsilon_0} \frac{e^2}{\left(\frac{4\pi\epsilon_0 \hbar^2}{m_e e^2}\right) n^2} $

$ E_n = -\frac{1}{8\pi\epsilon_0} \frac{e^2 m_e e^2}{4\pi\epsilon_0 \hbar^2 n^2} $

$ E_n = -\frac{m_e e^4}{(8\pi\epsilon_0)(4\pi\epsilon_0) \hbar^2 n^2} = -\frac{m_e e^4}{32\pi^2\epsilon_0^2 \hbar^2 n^2} $

Since $\hbar = h/(2\pi)$, $\hbar^2 = h^2/(4\pi^2)$.

$ E_n = -\frac{m_e e^4}{32\pi^2\epsilon_0^2 (h^2/4\pi^2) n^2} = -\frac{4\pi^2 m_e e^4}{32\pi^2\epsilon_0^2 h^2 n^2} $

$ E_n = -\frac{m_e e^4}{8\epsilon_0^2 h^2} \frac{1}{n^2} $

The term $\frac{m_e e^4}{8\epsilon_0^2 h^2}$ is a constant. Substituting the values of $m_e, e, \epsilon_0, h$:

$ \frac{m_e e^4}{8\epsilon_0^2 h^2} \approx 2.18 \times 10^{-18} \, J $

Converting this energy to electron volts: $1 \, J \approx 1/(1.602 \times 10^{-19}) \, eV$.

$ 2.18 \times 10^{-18} \, J \times \frac{1 \, eV}{1.602 \times 10^{-19} \, J} \approx 13.6 \, eV $

So, the energy levels of the hydrogen atom are quantised and given by:

$ E_n = -\frac{13.6}{n^2} \, eV $

Where $n=1, 2, 3, ...$ are the principal quantum numbers. The negative sign indicates that the electron is bound to the nucleus. The lowest energy level is for $n=1$ (ground state), $E_1 = -13.6 \, eV$. As $n$ increases, the energy becomes less negative (closer to zero), meaning the electron is less tightly bound. For $n \to \infty$, $E_\infty = 0$, which corresponds to the electron being free from the nucleus (ionisation). The ionisation energy of hydrogen is the energy required to remove the electron from the ground state ($n=1$) to $n=\infty$, which is $0 - (-13.6 \, eV) = 13.6 \, eV$.

Example 1. Calculate the energy of the electron in the second excited state of the hydrogen atom and express it in Joules.

Answer:

Given:

Hydrogen atom ($Z=1$).

The energy levels of the hydrogen atom are given by $E_n = -\frac{13.6}{n^2} \, eV$.

The ground state corresponds to $n=1$. The first excited state is $n=2$. The second excited state corresponds to $n=3$.

Substitute $n=3$ into the energy formula:

$ E_3 = -\frac{13.6}{3^2} \, eV = -\frac{13.6}{9} \, eV $

$ E_3 \approx -1.511 \, eV $

Now, convert the energy from electron volts to Joules. $1 \, eV = 1.602 \times 10^{-19} \, J$.

$ E_3 = -1.511 \, eV \times (1.602 \times 10^{-19} \, J/eV) $

$ E_3 \approx -2.419 \times 10^{-19} \, J $

The energy of the electron in the second excited state ($n=3$) of the hydrogen atom is approximately -1.511 eV or $-2.419 \times 10^{-19}$ Joules.


Radii of Orbits ($ r_n \propto n^2 $)

As derived from Bohr's postulates (Section I1), the radii of the stationary orbits in the hydrogen atom are quantised and given by the formula:

$ r_n = \left(\frac{4\pi\epsilon_0 \hbar^2}{m_e e^2}\right) n^2 $

Where $n=1, 2, 3, ...$ is the principal quantum number, $\epsilon_0$ is the permittivity of free space, $\hbar$ is the reduced Planck constant, $m_e$ is the electron mass, and $e$ is the elementary charge.

The term in the parenthesis is a constant, representing the radius of the first Bohr orbit ($n=1$), denoted as the Bohr radius ($a_0$).

$ a_0 = \frac{4\pi\epsilon_0 \hbar^2}{m_e e^2} \approx 0.0529 \, nm $

Thus, the radii of the allowed orbits are given by:

$ r_n = a_0 n^2 $

This shows that the radius of the $n$-th Bohr orbit is directly proportional to the square of the principal quantum number ($r_n \propto n^2$).

Allowed radii for hydrogen atom ($Z=1$):

The radii of the orbits increase rapidly as the quantum number $n$ increases. This means the electron is located farther from the nucleus in higher energy levels.

For hydrogen-like atoms (ions with only one electron, e.g., He$^+$, Li$^{++}$), the nucleus has a charge of $+Ze$. The derivation would involve replacing $e^2$ with $(Ze)e = Ze^2$ in the force equation. The radius formula becomes:

$ r_n = \frac{4\pi\epsilon_0 \hbar^2 n^2}{m_e (Ze^2)} = \frac{a_0 n^2}{Z} $

So, for a given $n$, the radius of the orbit in a hydrogen-like ion is $1/Z$ times the radius in a hydrogen atom. The orbits are smaller for ions with higher nuclear charge.

Example 1. Calculate the radius of the third Bohr orbit of the hydrogen atom.

Answer:

Given:

Hydrogen atom ($Z=1$). We need the radius of the third Bohr orbit, so $n=3$.

The radius of the $n$-th Bohr orbit of hydrogen is given by $r_n = a_0 n^2$, where $a_0 \approx 0.0529 \, nm$ is the Bohr radius.

Substitute $n=3$ and $a_0$:

$ r_3 = a_0 (3)^2 = 9 a_0 $

$ r_3 \approx 9 \times (0.0529 \, nm) $

$ r_3 \approx 0.4761 \, nm $

The radius of the third Bohr orbit of the hydrogen atom is approximately 0.4761 nanometers.

Example 2. Find the radius of the first Bohr orbit of the He$^+$ ion. ($Z=2$ for Helium).

Answer:

Given:

He$^+$ ion. This is a hydrogen-like ion with $Z=2$. We need the radius of the first orbit, so $n=1$.

The radius of the $n$-th Bohr orbit for a hydrogen-like ion is given by $r_n = \frac{a_0 n^2}{Z}$.

Substitute $n=1$, $Z=2$, and $a_0 \approx 0.0529 \, nm$:

$ r_1 = \frac{a_0 (1)^2}{2} = \frac{a_0}{2} $

$ r_1 \approx \frac{0.0529 \, nm}{2} = 0.02645 \, nm $

The radius of the first Bohr orbit of the He$^+$ ion is approximately 0.02645 nanometers. It is half the radius of the first Bohr orbit of the hydrogen atom due to the double nuclear charge attracting the electron more strongly.


Velocity of Electron in Orbits ($ v_n \propto 1/n $)

Using Bohr's postulates, we can also determine the allowed speeds of the electron in the stationary orbits of the hydrogen atom ($Z=1$).

Derivation of Velocity

From the second postulate (quantization of angular momentum), $m_e v_n r_n = n\hbar$. We found the expression for the radius $r_n = \frac{4\pi\epsilon_0 \hbar^2}{m_e e^2} n^2$.

We can solve the angular momentum equation for $v_n$:

$ v_n = \frac{n\hbar}{m_e r_n} $

Substitute the expression for $r_n$:

$ v_n = \frac{n\hbar}{m_e \left(\frac{4\pi\epsilon_0 \hbar^2}{m_e e^2}\right) n^2} $

$ v_n = \frac{n\hbar m_e e^2}{m_e 4\pi\epsilon_0 \hbar^2 n^2} $

Cancel out common terms ($m_e$, $n$, $\hbar$):

$ v_n = \frac{e^2}{4\pi\epsilon_0 \hbar n} $

This shows that the speed of the electron in the $n$-th Bohr orbit is proportional to $1/n$ ($v_n \propto 1/n$).

The speed of the electron in the first Bohr orbit ($n=1$) is $v_1$:

$ v_1 = \frac{e^2}{4\pi\epsilon_0 \hbar} $

Substitute the values of constants ($e = 1.602 \times 10^{-19} \, C$, $\epsilon_0 = 8.854 \times 10^{-12} \, C^2/(N \cdot m^2)$, $\hbar = 1.055 \times 10^{-34} \, J \cdot s$):

$ v_1 \approx 2.18 \times 10^6 \, m/s $

The speed of the electron in the first Bohr orbit is approximately $2.18 \times 10^6 \, m/s$, which is about 1/137 times the speed of light ($c \approx 3 \times 10^8 \, m/s$). This ratio $\frac{e^2}{4\pi\epsilon_0 \hbar c} = \frac{v_1}{c}$ is the fine-structure constant ($\alpha \approx 1/137$).

The speeds in the allowed orbits are $v_n = v_1 / n$. As $n$ increases, the electron's speed decreases.

For hydrogen-like atoms (ions with charge $+Ze$), the formula for velocity becomes:

$ v_n = \frac{Ze^2}{4\pi\epsilon_0 \hbar n} = \frac{Z v_1}{n} $

The speed is proportional to $Z/n$. For a given $n$, the electron moves faster in ions with higher nuclear charge.

Example 1. Calculate the speed of the electron in the second Bohr orbit of the hydrogen atom.

Answer:

Given:

Hydrogen atom ($Z=1$). We need the speed in the second Bohr orbit, so $n=2$.

The speed of the electron in the $n$-th Bohr orbit of hydrogen is given by $v_n = v_1 / n$, where $v_1 \approx 2.18 \times 10^6 \, m/s$ is the speed in the first Bohr orbit.

Substitute $n=2$ and $v_1$:

$ v_2 = \frac{v_1}{2} $

$ v_2 \approx \frac{2.18 \times 10^6 \, m/s}{2} = 1.09 \times 10^6 \, m/s $

The speed of the electron in the second Bohr orbit of the hydrogen atom is approximately $1.09 \times 10^6 \, m/s$. It is half the speed in the ground state.

Example 2. Find the speed of the electron in the first Bohr orbit of the Li$^{++}$ ion. ($Z=3$ for Lithium).

Answer:

Given:

Li$^{++}$ ion. This is a hydrogen-like ion with $Z=3$. We need the speed in the first orbit, so $n=1$.

The speed of the electron in the $n$-th Bohr orbit for a hydrogen-like ion is given by $v_n = \frac{Z v_1}{n}$, where $v_1 \approx 2.18 \times 10^6 \, m/s$ is the speed in the first Bohr orbit of hydrogen.

Substitute $n=1$, $Z=3$, and $v_1$:

$ v_1 = \frac{3 \times v_1}{1} = 3 v_1 $

$ v_1 \approx 3 \times (2.18 \times 10^6 \, m/s) = 6.54 \times 10^6 \, m/s $

The speed of the electron in the first Bohr orbit of the Li$^{++}$ ion is approximately $6.54 \times 10^6 \, m/s$. It is three times the speed in the first orbit of hydrogen due to the triple nuclear charge.