Bohrs Model - History, Structure, Theories, Limitations, FAQs

Postulates of Bohr Model Atom

• In an atom, electrons (badly charged) revolve around a well-charged nucleus in a clear circular pattern called orbits or shells.
• Each circle or shell has a fixed force and these circular lines are known as orbital shells.
• Power levels are represented by a whole number (n = 1, 2, 3…) known as a quantum number.
• This quantum numerical range starts from the side of the nucleus with n = 1 having the lowest energy level.
• Orbits n = 1, 2, 3, 4, and so on... are assigned as K, L, M, N, and so on… shells, and when the electron reaches a very low level of energy, it is said to be in the ground state.
• Electrons in the atom go from low energy to high energy by getting the energy needed and electrons move from high energy to high energy levels by losing energy.

Limitations of Bohr Model Atom

• Bohr's atomic model failed to explain the effect of Zeeman (the effect of magnetic force on a series of atoms).
• It also failed to explain the effect of Stark (the effect of the electric field on the atomic spectrum).
• It violates the Heisenberg Uncertainty Principle rule.
• Could not define spectra obtained from large atoms.

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History of the Bohr Model

The discovery of electrons and radioactivity in the late 19th century led to a variety of proposed atomic formations.

In 1913, Niels Bohr proposed the idea of a hydrogen atom, based on the quantum assumption that a certain body part takes only different amounts. Electrons revolve around the nucleus, but only in fixed lines, and When electrons cross the low-energy path, the difference is sent as radiation. Bohr's model explained why atoms only emit light at fixed wavelengths, and later inject ideas into light quanta.

Modified Bohr Model

Despite the success of the Bohr model, there have been major flaws. On the test side, a detailed analysis of hydrogen extraction found a single extraction line actually consisting of two or more closely spaced lines, a feature not found in the Bohr model. Theoretically, the Bohr model mixes particle images with electron waves, which were considered by many to be unsatisfactory.

For these reasons, better treatment of hydrogen atoms was bought when the electron was considered a wave from the beginning. This theory, developed by Heisenberg, Pauli, Schrödinger, Sommerfeld, and others, is well mathematically accurate. The main result from this theory is that four Quantum Numbers describe the state of the electron, compared to the single quantum n number present in the Bohr model. These quantum numbers are as follows:

• n, the principle quantum number. This is similar to the Bohr type, and can take up values of 1,2,3, ....
• l, orbital quantum number. This is a label that describes the magnitude of the angular pressure of an electron. With a given n, l can take values 0,1,2, ..., n - 1.
• m_l, the number of spin-orbital quantum. This is a label that describes the angular component of an electron vector. For a given l, ml can take values - l, - l + 1, ..., - 1,0,1, ..., l - 1, l.
• m_s, the quantum spin number. This label, in a very limited sense, can be regarded as setting a point where an electron rotates on its axis. ms can take one of two values, $ \ pm $ 1/2.

Therefore, for a given n, there may be 2n² areas with different values of l, m_l, and m_s.

Historically, the primary quantum n label is called a Shell, and n = 1,2,3, ... the shell is sometimes called the K, L, M, ... shell.

The number of orbital quantum l labels subshells, and l = 0,1,2,3,4, ... the subshell is also called s, p, d, f, ... subshell.

If we consider starting electrons to form different atoms, then one would expect all the electrons to enter the lowest energy state, namely n = 1 and l = 0 = ml. This does not happen in nature, however. Paul explained this by posting that the electrons adhere to what is now called the Pauli emission system:

No two electrons in a system can have the same sets of quantum numbers.

Some Solved Examples

Question 1: The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to $\frac{h^2}{x m a_0^2}$. The value of $2 x$ is
1) 422
2) 528
3) (correct) 631
4) 122

Solution:

$\begin{aligned}
& \frac{h^2}{x m a_0^2} \\
& x=\frac{8 \pi^2 \times 16}{4}=32 \pi^2 \\
& x=32(3.14)^2 \\
& x=315.5 \\
& 2 x=2 \times 315.5 \\
& =631
\end{aligned}$
Hence, the answer is the option (3).

Question 2: The radius of a gold nucleus is $7.24 \times 10^{-15} \mathrm{~m}$. Calculate the volume of the nucleus.

1) $1.15 \times 10^{-44} \mathrm{~m}^3$

2) (correct) $2.81 \times 10^{-44} m^3$

3)$1.52 \times 10^{-44} \mathrm{~m}^3$

4) $3.45 \times 10^{-44} \mathrm{~m}^3$

Solution:

The volume of a sphere is given by the formula $V=\frac{4}{3} \pi r^3$, where $r$, where r is the radius of the sphere. Substituting the given value of the radius, we get: $V=\frac{4}{3} \times \pi \times\left(7.24 \times 10^{-15}\right)^3=2.81 \times 10^{-44} \mathrm{~m}^3$
Therefore, the volume of the gold nucleus is $2.81 \times 10^{-44} \mathrm{~m}^3$.
Hence, the answer is the option (1).

Question 3: The ratio of the energy of the electron in the ground state of hydrogen to the electron in the first excited state of $B e^{3+}$ is
1) (correct) 1:4
2) 1:8
3) 1:16
4) 16:1

Solution:

As we learnt in
Total energy of electron in nth orbit -

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

Where $z$ is atomic number
We know that total energy of electron in nth orbit, $E_n=-13.6 \frac{z^2}{n^2} \mathrm{eV}$

$E_1(\text { ground state of hydrogen })=-13.6 \times \frac{1}{1} \mathrm{eV}$

$\mathrm{E}_2$ (first excited state of $\mathrm{Be}^{3+}$ ) $=-13.6 \times \frac{4 \times 4}{2 \times 2}$

$(n=2, z=4)$

$=-13.6 \times 4 \mathrm{eV}$
$\therefore E_1: E_2=1: 4$
Hence, the correct answer is option (1)

Question 4: If the number of revolutions made by electron in 1.0 s in H atom in its $n^{\text {th }}$ orbit is twice of the number of revolutions made by electron in 1.0 s in the $2^{\text {nd }}$ orbit of $\mathrm{He}^{+}$ion atom, then $n$ is
1) (correct) 01
2) 02
3) 03
4) 04

Solution:

(a)

$\begin{aligned}
& \text { no. of revolution }=\frac{V}{2 \pi r}=\frac{v_0 z}{n \times 2 \pi \frac{n^2 a_0}{z}}=N \\
& \Rightarrow N=\frac{V_0}{2 \pi a_0}\left(\frac{z^2}{n^3}\right)=\frac{k z^2}{n^3}
\end{aligned}$

For $\mathrm{He}^{+}$in 2 nd orbit $\mathrm{N}_2=\frac{\mathrm{k}(2)^2}{2^3}=\frac{k}{2}$

$\begin{aligned}
& \Rightarrow 2 N_2=k=N_1 \text { (no of revolutions in H-atom) } \\
& \Rightarrow N_1=k=\frac{k}{n^3} \Rightarrow n=1
\end{aligned}$

Hence, the answer is the option (1).

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