Question

When is ${H_\alpha }$line in the emission spectrum of a hydrogen atom obtained? Calculate the frequency of the photon emitted during this transition.

Answer 1

Hint: H-alpha is a spectral line which is deep-red. This can be seen in the Balmer series for the shortest transition. For the frequency, we will first find the
corresponding energy in the 3rd orbit 2nd orbit and then find the difference. We will use the
Planck’s formula which relates energy and the frequency to find the answer.

Formula used:
The formula which relates energy and the frequency is given below:
$E = hf$
Where, $h$ is the Planck’s constant, $f$ is the frequency, $E$ is the energy.

Complete step by step solution:
There are six series in hydrogen spectrum they are
Lyman series, Balmer series, Paschen series, Brackett series, Pfund series and Humphreys series. Out of these six only Balmer series is imaginatively called as alpha, beta, gamma and so on…

so ${H_\alpha }$ electron transition in hydrogen atom must take place 3rd orbital to 2nd orbital
i.e; electron jumps from n=3 to n=2.
For hydrogen atom energy in ${n^{th}}$orbital
${E_n} = - {{13.6}}{{{n^2}}}eV$
So, for n=3
$
{E_3} = - {{13.6}}{{{3^2}}}eV \\
\Rightarrow {E_3} = - 1.51eV \\ $
And for n=2
$
{E_2} = - {{13.6}}{{{2^2}}}eV \\
\Rightarrow {E_2} = - 3.4eV \\ $
Energy difference between these two shells will be
$
\Delta E = {E_3} - {E_2} \\
\Rightarrow \Delta E = - 1.51 - ( - 3.4) \\
\Rightarrow \Delta E = 1.89eV \\ $
This will be the energy of the emitted photon.
Using planck's quantum theory we have
$E = hf$
$
\Rightarrow f = {E}{h} \\
\Rightarrow f = {{1.89 \times 1.6 \times {{10}^{ - 19}}}}{{6.63 \times 10 \times {{10}^{ - 34}}}} \\
\therefore f = 4.56 \times {10^{14}}Hz \\ $
Hence frequency of ${H_\alpha }$is $4.56 \times {10^{14}}Hz$.

Note:The Balmer series lines that we see are imaginatively called alpha, beta, gamma and so on…, so ${H_\alpha }$ line corresponds to the first line of the Balmer series in other words ${H_\alpha }$ spectrum is produced when electrons from a hydrogen atom jumps from 3rd orbital to 2nd orbital as for Balmer series lowest orbital is 2nd orbital. It is important to remember that higher the energy emitted by the electron, higher will be its frequency.