Bohr Model Of The Hydrogen Atom

The Bohr model of the hydrogen atom represents a groundbreaking advancement in atomic theory, providing a clear and quantifiable explanation of atomic structure. Proposed by Niels Bohr in 1913, this model introduces the concept that electrons orbit the nucleus in discrete energy levels or "shells," rather than in continuous orbits. This revolutionary idea helped explain the discrete lines observed in atomic spectra, known as spectral lines. In real life, the Bohr model's principles underpin technologies such as lasers, which rely on electron transitions between energy levels, and various spectroscopy techniques used in chemical analysis and astronomy. By understanding the Bohr model, we gain insights into the fundamental workings of atoms, influencing both scientific research and practical applications. In this article, we will explore the Bohr model of the hydrogen atom, its key concepts, and its implications for modern science and technology.

Bohr's Model of Hydrogen Atom

Bohr proposed a model for the hydrogen atom which is also applicable to some lighter atoms in which a single electron revolves around a stationary nucleus of positive charge $Z_e$ (called a hydrogen-like atom)
Bohr's model is based on the following postulates

(1). Bohr’s first postulate was that an electron in an atom could revolve in certain stable orbits without the emission of radiant energy, contrary to the predictions of electromagnetic theory. According to this postulate, each atom has certain definite stable states in which it can exist, and each possible state has definite total energy. These are called the stationary states of the atom

For electrons revolving in a stable orbit, the necessary centripetal force is provided by the coulomb's force

$\frac{m v_n^2}{r_n}=\frac{k z e^2}{r_n^2}$
(2). Bohr’s second postulate defines these stable orbits. This postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of $\frac{h}{2 \pi}$ where h is the Planck’s constant (= 6.6 × 10^–34 J s). Thus the angular momentum (L) of the orbiting electron is quantised. That is $L=m v_n r_n=\frac{n h}{2 \pi} ; n=1,2,3 \ldots \ldots \infty$
(3). Bohr’s third postulate incorporated into atomic theory the early quantum concepts that had been developed by Planck and Einstein. It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is then given by

E_i is the energy of the initial state and E_f is the energy of the final state. Also, E_i > E_f.

r_n-radius of the nth orbit

v_n- speed of an electron in the nth orbit

Radius of Orbit and Velocity of the Electron

In the Bohr model of the hydrogen atom, the radius of the electron's orbit and its velocity are crucial parameters that help define the atom's structure. According to Bohr's theory, electrons orbit the nucleus in quantized orbits, and each orbit corresponds to a specific energy level.

The Radius of the Orbit

The radius of an electron's orbit in the Bohr model of the hydrogen atom is a key concept in understanding atomic structure. According to Bohr's theory, electrons orbit the nucleus in quantized paths or orbits with specific radii. The radius of the electron's orbit depends on the principal quantum number n. For an electron around a stationary nucleus, the electrostatic force of attraction provides the necessary centripetal force.

ie. $\frac{1}{4 \pi \varepsilon_0} \frac{(Z e) e}{r^2}=\frac{m v^2}{r}$
also $m v r=\frac{n h}{2 \pi}$
From equations (i) and (ii) radius of the $r$ orbit
$
\begin{aligned}
& r_n=\frac{n^2 h^2}{4 \pi^2 k Z m e^2}=\frac{n^2 h^2 \varepsilon_0}{\pi m Z e^2}=0.53 \frac{n^2}{Z} \ \quad\left(k=\frac{1}{4 \pi \varepsilon_0}\right) \\
& \Rightarrow r_n \propto \frac{n^2}{Z} \\
\Longrightarrow & r_n=0.53 \frac{n^2}{Z} \
\end{aligned}
$

Speed of Electron

In the Bohr model of the hydrogen atom, the speed of the electron in its orbit is a crucial aspect of understanding atomic structure. According to Bohr's theory, From the above relations, the speed of electrons in $n^{t h}$ orbit can be calculated as

$
v_n=\frac{2 \pi k Z e^2}{n h}=\frac{Z e^2}{2 \varepsilon_0 n h}=\left(\frac{c}{137}\right) \frac{Z}{n}=2.2 \times 10^6 \frac{Z}{n} m / \mathrm{sec}
$
where $\left(c=\right.$ speed of light $\left.=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)$

Solved Examples Based on the Bohr Model of the Hydrogen Atom

Example 1: According to Bohr’s theory, the time average magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the n^th orbit is proportional to : (n=principal quantum number)

1) n⁻⁴

2) n⁻⁵

3) n⁻³

4) n⁻²

Solution:

Magnetic field due to current at the centre of the circle $=\frac{\mu_0 I}{2 r}$

$\begin{aligned} & I=\frac{q}{t}=\frac{e}{\left(\frac{2 \pi r}{v}\right)}=\frac{e v}{2 \pi r} \\ & B=\frac{\mu \cdot\left(\frac{e v}{2 \pi r}\right)}{2 r}=\frac{\mu_0 e v}{4 \pi r^2} \\ & B=\frac{\left(\frac{\mu e e}{4 \pi}\right) \cdot\left(\frac{c}{137}\right) \frac{z}{n}}{\left(r_0 \frac{n^2}{z}\right)^2}=\frac{\mu_0 e c}{4 \pi \times 137 r_0^2} \times \frac{z^3}{n^5} \\ & B \propto n^{-5}\end{aligned}$

Hence, the answer is the option (2).

Example 2: An electron from various excited states of a hydrogen atom emits radiation to come to the ground state. Let λ_n, and λ_g be the de Broglie wavelength of the electron in the n^th state and the ground state respectively. Let $\lambda_n$ be the wavelength of the emitted photon in the transition from the nth state to the ground state. For large n (A, B are constants)

1) $\Lambda_n^2 \approx \lambda$
2) $\Lambda_n \approx A+\frac{B}{\lambda_n^2}$
3) $\Lambda_n \approx A+B \lambda_n$
4) $\Lambda_n^2 \approx A+B \lambda_n^2$

Solution:

The velocity of the electron in the nth orbital

$\begin{aligned} & v=\left(\frac{e^2}{2 \epsilon_0 h}\right) \frac{z}{n} \\ & \text { wherein } \\ & v \alpha \frac{z}{n} \\ & \frac{e^2}{2 \epsilon_0 h}=\frac{c}{137} \\ & V^n=\left(\frac{e^2}{2 \epsilon_0 h}\right) \cdot \frac{1}{n} \\ & E^n=\emptyset-\frac{1}{2} m V_n^2 \\ & \lambda_n=\frac{h}{m V_n}=\frac{h}{m \cdot \frac{e^2}{2 \epsilon_0 n h}}=\frac{2 \epsilon_0 n h^2}{m e^2}\left(\lambda_{n>>>\lambda_g}\right)\end{aligned}$

$\begin{aligned} & \frac{h c}{\Lambda_n}=E_n-E_1=\frac{1}{2} m\left(v_1^2-v_n^2\right)=\frac{1}{2} m\left(\left(\frac{h^2}{m \lambda g}\right)-\left(\frac{h}{m \lambda n}\right)\right) \\ & \frac{h c}{\Lambda_n}=\frac{h^2}{2 m}\left(\frac{1}{\lambda_g^2}-\frac{1}{\lambda_n^2}\right)=\frac{h^2}{2 m \lambda_g^2}\left(1-\frac{\lambda_g^2}{\lambda_n^2}\right) \\ & \frac{\Lambda_n^0}{h c}=\frac{2 m \lambda_g^2}{h^2} \cdot \frac{1}{1-\frac{\lambda_2^2}{\lambda_n^2}} \\ & \because \frac{\lambda_g}{\lambda_n}<<1 \text { so using Binomial expansion } \\ & \Lambda_n=\frac{2 m c \lambda_g^2}{h}\left(1+\frac{\lambda_g^2}{\lambda_n^2}\right) \Rightarrow \Lambda_n=A+\frac{B}{\lambda_n^2}\end{aligned}$

Hence, the answer is the option (2).

Example 3: The ratio of radii of the first three orbits in a hydrogen atom is:

1) 1: 4: 9

2) 1: 2: 3

3) 1: 2: 4

4) 9: 4: 1

Solution:

Atomic number, $Z$ is equal to 1
Hence the radius of $n^{\text {th }}$ orbit, $r_n=0.529 n^2 A^0$.
For the first three orbits, n values are 1,2 and 3.
Therefore:
The ratio of radii of first three orbits $=r_1: r_2: r_3=n_1^2: n_2^2: n_3^2=1^2: 2^2: 3^2=1: 4: 9$

Hence, the answer is the option (1).

Example 4: The time period of revolution of an 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) $5.6 \times 10^{12}$
2) $1.6 \times 10^{14}$
3) $7.8 \times 10^{14}$
4) $6.2 \times 10^{15}$

Solution:

The velocity of the n_th orbit

$
V_n \propto \frac{z}{n}
$
the radius of the $n_{\text {th }}$ orbit
$
\begin{aligned}
& r_n \alpha \frac{n^2}{z} \\
& T=\frac{2 \pi r}{V} \\
& \Rightarrow T \propto \frac{n^2 \times n}{z \times z}=\frac{n^3}{z^2} \\
& \Rightarrow T \propto \frac{1}{f} \\
& \Rightarrow f \propto \frac{z^2}{n^3} \\
& f_1=\frac{1}{1.6 \times 10^{-16}} s^{-1}
\end{aligned}
$
and
$
\begin{aligned}
& \frac{f_1}{f_2}=\left(\frac{n_2}{n_1}\right)^3=2^3=8 \\
\Rightarrow & f_2=\frac{f_1}{8}=7.8 \times 10^{14} \mathrm{~s}^{-1}
\end{aligned}
$

Hence, the answer is the option (3).

Example 5: In Bohr's model of a hydrogen-like atom, the force between the nucleus and the electrons is modified as $F=\frac{e^2}{4 \pi \epsilon_0}\left(\frac{1}{r^2}+\frac{\beta}{r^3}\right)$ Where $\beta$ is a constant. For the atom, the radius of the nth orbit in terms of the Bohr radius

$\left(a_0=\frac{\epsilon_0 h^2}{m \pi e^2}\right)$

1) $r_n=a_0 n-\beta$
2) $r_n=a_0 n^2+\beta$
3) $r_n=a_0 n^2-\beta$
4) $r_n=a_0 n+\beta$

Solution:

$
\begin{aligned}
& F=\frac{m v^2}{r}=\frac{e^2}{4 \pi \varepsilon_0}\left(\frac{1}{r^3}+\frac{\beta}{r^3}\right) \\
& \therefore v^2=\frac{e^2}{4 \pi \varepsilon_0 m}\left(\frac{1}{r}+\frac{\beta}{r^2}\right)
\end{aligned}
$

From Bohr's postulate
$
\begin{aligned}
& m v r=\frac{n h}{2 \pi} \\
& \therefore v=\frac{n h}{2 \pi m r}
\end{aligned}
$
comparing both we get
$
\frac{n^2 h^2}{4 \pi^2 m^2 r^2}=\frac{e^2}{4 \pi \varepsilon_0 m}\left(\frac{1}{r}+\frac{\beta}{r^2}\right)
$

So,
$
r_n=a_0 n^2-\beta
$

Hence, the answer is the option (3).

Summary

The Bohr model of the hydrogen atom introduced significant insights into atomic structure by proposing that electrons orbit the nucleus in discrete, quantized orbits. Bohr's theory quantifies the radius of these orbits and the speed of electrons, with the radius increasing with the principal quantum number and the speed decreasing in higher orbits. This model explains the stability of electron orbits and the emission of photons during transitions between these orbits. The principles derived from Bohr's model are foundational in understanding atomic spectra and underpin technologies like lasers and spectrometers, demonstrating their relevance to both scientific and practical applications.