Question

Using Bohr’s postulates derive the expression for frequency of radiation submitted when electron in hydrogen atom undergoes transition from higher energy state (quantum number ${n_i}$ ) to the lower state, (${n_f}$) when electron in hydrogen atom jumps from energy state ${n_i} = 4$ to ${n_f} = 3,2,1$. Identify the special series to which the emission lines holding belong.

Answer 1

To find the solution of the given question, first we need to know the formula of De Broglie. The Bohr model consists of a small positively charged nucleus orbited by a negatively charged nucleus. In Bohr’s theory the hydrogen atom is based on the quantum theory that energy is transferred in certain quantities.

Formula used:
De Broglie wavelength,
$\lambda = \dfrac{h}{{mv}}$
Where, $\lambda$ is wavelength,
$h$ is Planck's constant,
$m$ is mass and
$v$ is velocity

Complete step by step solution:
We know that the kinetic energy is equal to the electrostatic energy by the law of the energy conservation, only when the charged particle is accelerated by potential $V$.
$\Rightarrow$ $\dfrac{1}{2}m{v^2} = q{v^{}}$
$\Rightarrow$ $v = \dfrac{{\sqrt {2qv} }}{m}$
Therefore the De Broglie wavelength,
$\Rightarrow$ $\lambda = \dfrac{h}{p}$
$\Rightarrow$ $\lambda = \dfrac{h}{{mv}}$
$\Rightarrow$ $\lambda = \dfrac{h}{{\sqrt {2mqv} }}$
In the hydrogen atom the radius of an electron orbit,
$\Rightarrow$ $r = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}km{e^2}}}$
The kinetic energy of an electron is,
$\Rightarrow$ ${E_k} = \dfrac{1}{2}m{v^2}$
$\Rightarrow$ ${E_k} = \dfrac{{k{e^2}}}{r}$
Using the formula of radius of electron orbit we get,
$\Rightarrow$ ${E_k} = \dfrac{{h{e^2}4{\pi ^2}km{e^2}}}{{{n^2}{h^2}}}$
$\Rightarrow$ ${E_k} = \dfrac{{2{\pi ^2}{k^2}m{e^4}}}{{{n^2}{h^2}}}$
We need to find the potential energy,
The potential energy is,
$\Rightarrow$ ${E_k} = \dfrac{{ - k(e) \times (e)}}{r}$
$\Rightarrow$ ${E_k} = \dfrac{{ - k{e^2}}}{r}$
Using the formula of radius of an electron orbit we get,
$\Rightarrow$ ${E_p} = - k{e^2} \times \dfrac{{4{\pi ^2}km{e^2}}}{{{n^2}{h^2}}}$
$\Rightarrow$ ${E_p} = - {\dfrac{{2{\pi ^2}{k^2}m{e^4}}}{{{n^2}{h^2}}}^{}}$
$\Rightarrow$ ${E_p} = \dfrac{{2{\pi ^2}{k^2}m{e^4}}}{{{h^2}}} \times \left[ {\dfrac{1}{{{n^2}}}} \right]$
According to Bohr’s frequency condition, the electron in the hydrogen atom undergoes transition from the higher energy state to the lower energy state (${n_f}$) is,
$\Rightarrow$ $hv = {E_m} - {E_{nf}}_{}$
$\Rightarrow$ $hv = - \dfrac{{2{\pi ^2}{k^2}m{e^4}}}{{{h^2}}} \times \dfrac{1}{{n_i^2}} - \left[ {\dfrac{{ - 2{\pi ^2}{k^2}m{e^4}}}{{{h^2}}} \times \dfrac{1}{{n_i^2}}} \right]$
$\Rightarrow$ $hv = \dfrac{{2{\pi ^2}{k^2}m{e^4}}}{{{h^2}}} \times \left[ {\dfrac{1}{{n_i^2}} - \dfrac{1}{{n_i^2}}} \right]$
$\Rightarrow$ $v = \dfrac{{2{\pi ^2}{k^2}m{e^4}}}{{{h^3}}} \times \left[ {\dfrac{1}{{n_i^2}} - \dfrac{1}{{n_i^2}}} \right]$
$\Rightarrow$ $v = \dfrac{{c2{\pi ^2}{k^2}m{e^4}}}{{c{h^3}}} \times \left[ {\dfrac{1}{{n_i^2}} - \dfrac{1}{{n_i^2}}} \right]$
$\Rightarrow$ $\dfrac{{2{\pi ^2}{k^2}m{e^4}}}{{c{h^3}}} = R =$Rydberg constant
$\Rightarrow$ $R = 1.097 \times {10^7}m - 1$
Thus,
$\Rightarrow$ $v = Rc \times \left[ {\dfrac{1}{{n_i^2}} - \dfrac{1}{{n_i^2}}} \right]$
So the higher state is ${n_i} = 4$
And the lower state is ${n_f} = 3,2,1$.
The transition is,
If ${n_i} = 4$ to ${n_f} = 3$, then it is a Paschen series
If ${n_i} = 4$ to ${n_f} = 2$, then it is a Balmer Series
If ${n_i} = 4$ to ${n_f} = 1$, then it is a Lyman Series

Note: In Bohr’s postulates the hydrogen atom the electron moves in the circular orbit and the angular momentum of an electron in the orbit is quantized and also change of electron energy takes place which jumps from one orbit to another.