Question

what is the colour of the second line of the Balmer series?
\[(a).\] blue-green
\[(b).\] yellow
\[(c).\] red
\[(d).\] blue

Answer 1

Hint: The second line of Balmer series will be obtained when an electron comes back from \[N\] shell to k shell which means from ${{n}_{4}}$ to ${{n}_{2}}$ orbit.

Complete step by step answer:
The colour of the second line of the Balmer series can be observed by comparing the energy and wavelength terms. wavelength $(\lambda )$ is defined as the distance between two consecutive crest and trough. It is measured in $m,cm,nm,{{A}^{\circ }}$ etc. we know that energy is inversely proportional to wavelength therefore higher the energy lower is the wavelength and thereby by observing in which region the following wavelength falls we can find out the colour .
\[\Delta E=\dfrac{hc}{\lambda }\] where $h$ is Planck’s constant , $c$ is speed of light , $\Delta E$ is difference in energy and \[\lambda \]is wavelength.
Now we will first calculate the difference in energy using the formula
$\Delta E=-13.6\times {{Z}^{2}}\{\dfrac{1}{n_{2}^{2}}-\dfrac{1}{n_{1}^{2}}\}eV$ Where $Z$ is atomic number, ${{n}_{2}}=4$ and ${{n}_{1}}=2$
Thus putting all the values in above formula we get,$\Delta E=-13.6\times {{1}^{2}}\{\dfrac{1}{{{4}^{2}}}-\dfrac{1}{{{2}^{2}}}\}$
$\Rightarrow $\[\Delta E=-13.6\{\dfrac{1}{16}-\dfrac{1}{4}\}eV\]
$\Rightarrow $$\Delta E=-13.6\{\dfrac{1-4}{16}\}eV$
$\Rightarrow $$\Delta E=-13.6\{\dfrac{-3}{16}\}eV$
$\Rightarrow 2.55eV$$=2.55\times 1.602\times {{10}^{-19}}J$
$\Rightarrow $$\Delta E=4.0851\times {{10}^{-19}}J$
Now putting the value of $\Delta E$ in \[\Delta E=\dfrac{hc}{\lambda }\] we can find out the value of $\lambda $
$\Rightarrow $$\lambda =\dfrac{hc}{\Delta E}$ $=\dfrac{6.626\times {{10}^{-34}}Js\times 3\times {{10}^{8}}m{{s}^{-1}}}{4.0851\times {{10}^{-19}}J}$
$\Rightarrow $=$\lambda =4.86\times {{10}^{-34+8+19}}$ $=4.865\times {{10}^{-7}}m$
$\Rightarrow $ $\lambda =4865{{A}^{\circ }}$
This value of wavelength belongs to blue colour. So the correct option is \[d\].
First line of any series \[=\alpha \]
Second line of any series $=\beta $
Third line of any series $=\gamma $

Therefore, the correct answer is option D.

Note: Balmer series occurs in visible region .Apart from Balmer series ,we have five more series named as Lyman (U.V region), Paschen (Infrared), Brackett (I.R. region), Pfund (I.R region) and Humphery (Far I.R region).