Question 13 Marks
On the basis of Bohr's postulate obtain the formula for radius and total energy of electron in the $n^{\text {th }}$ stable orbit for hydrogen atom.
Answer
View full question & answer→→ Bohr combined classical and early quantum concepts and gave his theory in the form of three postulates. These are :
→ (i) Bohr's first postulate : An electron in an atom could revolve in certain stable orbits without the emission of radiant energy.
→ 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.
→This contrary to the predictions of electromagnetic theory.
→ (ii) Bohr's second postulate : The electron revolves around the nucleus only in those orbits for which the angular momentum is in integral multiple of $\frac{h}{2 \pi}$.
→ Where, $h$ is Planck's constant
$
\begin{array}{l}
h=6.625 \times 10^{-34} J s . \\
L=\frac{n h}{2 \pi} \text { Where, } n=1,2,3 \ldots .
\end{array}
$
→ (iii) Bohr's third postulate : An electron makes 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
$
h v=E_i-E_f
$
Where $E _i$ and $E _f$ are the energies of the initial and final states and $E _i> E _f$.
→ From Bohr's second postulate, the formula for the radius of $n^{\text {th }}$ orbit for hydrogen atom is.
$
r_n=\frac{n^2 h^2 \varepsilon_0}{\pi m e^2} \ldots(1)
$
→ The total energy of the electron in the stationary states of the hydrogen atom is
$
E_n=-\frac{1}{8 \pi \varepsilon_0} \frac{e^2}{r_n} \ldots(2)
$
→ Using equation (1) and equation (2)
$
\begin{array}{l}
E_n=-\frac{1}{8 \pi \varepsilon_0} \frac{e^2}{\left(\frac{n^2 h^2 \varepsilon_0}{\pi m e^2}\right)} \\
\therefore E_n=-\frac{m e^4}{8 \varepsilon_0^2 n^2 h^2}
\end{array}
$
→ Substituting $m=9.1 \times 10^{-31} kg$
$
\begin{array}{l}
e=1.6 \times 10^{-19} C \\
\varepsilon_0=8.85 \times 10^{-12} \frac{C^2}{Nm^2} \\
h=6.625 \times 10^{-34} Js
\end{array}
$
→ Simplifying equation,
$
E_n=-\frac{2.18 \times 10^{-18}}{n^2} J
$
→ Atomic energies are often expressed in electron volts $( eV )$.
$
\begin{array}{l}
\therefore E_n=-\frac{2.18 \times 10^{-18}}{n^2 \times 1.6 \times 10^{-19}} eV \\
\therefore E_n=-\frac{13.6}{n^2} eV
\end{array}
$
→ The negative sign of the total energy of an electron moving in an orbit means that the electron is bound with the nucleus.
→ Thus, energy will be required to remove the electron from the hydrogen atom to a distance infinitely far away from its nucleus.
→ (i) Bohr's first postulate : An electron in an atom could revolve in certain stable orbits without the emission of radiant energy.
→ 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.
→This contrary to the predictions of electromagnetic theory.
→ (ii) Bohr's second postulate : The electron revolves around the nucleus only in those orbits for which the angular momentum is in integral multiple of $\frac{h}{2 \pi}$.
→ Where, $h$ is Planck's constant
$
\begin{array}{l}
h=6.625 \times 10^{-34} J s . \\
L=\frac{n h}{2 \pi} \text { Where, } n=1,2,3 \ldots .
\end{array}
$
→ (iii) Bohr's third postulate : An electron makes 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
$
h v=E_i-E_f
$
Where $E _i$ and $E _f$ are the energies of the initial and final states and $E _i> E _f$.
→ From Bohr's second postulate, the formula for the radius of $n^{\text {th }}$ orbit for hydrogen atom is.
$
r_n=\frac{n^2 h^2 \varepsilon_0}{\pi m e^2} \ldots(1)
$
→ The total energy of the electron in the stationary states of the hydrogen atom is
$
E_n=-\frac{1}{8 \pi \varepsilon_0} \frac{e^2}{r_n} \ldots(2)
$
→ Using equation (1) and equation (2)
$
\begin{array}{l}
E_n=-\frac{1}{8 \pi \varepsilon_0} \frac{e^2}{\left(\frac{n^2 h^2 \varepsilon_0}{\pi m e^2}\right)} \\
\therefore E_n=-\frac{m e^4}{8 \varepsilon_0^2 n^2 h^2}
\end{array}
$
→ Substituting $m=9.1 \times 10^{-31} kg$
$
\begin{array}{l}
e=1.6 \times 10^{-19} C \\
\varepsilon_0=8.85 \times 10^{-12} \frac{C^2}{Nm^2} \\
h=6.625 \times 10^{-34} Js
\end{array}
$
→ Simplifying equation,
$
E_n=-\frac{2.18 \times 10^{-18}}{n^2} J
$
→ Atomic energies are often expressed in electron volts $( eV )$.
$
\begin{array}{l}
\therefore E_n=-\frac{2.18 \times 10^{-18}}{n^2 \times 1.6 \times 10^{-19}} eV \\
\therefore E_n=-\frac{13.6}{n^2} eV
\end{array}
$
→ The negative sign of the total energy of an electron moving in an orbit means that the electron is bound with the nucleus.
→ Thus, energy will be required to remove the electron from the hydrogen atom to a distance infinitely far away from its nucleus.
