Question

Taking the wavelength of first Balmer line in hydrogen spectrum ($n=3$ to $n=2$) as $660nm$, the wavelength of the second Balmer line($n=4$ to $n=2$)will be:
$A.889.2nm$
$B.642.7nm$
$C.488.9nm$
$D.388.9nm$

Answer 1

- In the analysis of lines spectra of hydrogen atom Lyman, Balmer, Paschen, Brackett and Pfund series are used to show the energy change during transition of electrons from one energy level to another in a hydrogen atom. The Balmer series represents the hydrogen spectrum when the initial quantum number is 2 and the final quantum numbers may be 3, 4, 5………. The Balmer series is applicable for hydrogen atoms only.
Formula used:
For wavelength of a Balmer line we use the following relation:-
$\dfrac{1}{\lambda }=R\left( \dfrac{1}{n_{i}^{2}}-\dfrac{1}{n_{f}^{2}} \right)$

Complete answer:
We are going to find wavelength of second Balmer line by using the following relation:-
$\dfrac{1}{\lambda }=R\left( \dfrac{1}{n_{i}^{2}}-\dfrac{1}{n_{f}^{2}} \right)$Where symbols have their meaning listed below
$\lambda =$Wavelength of radiation
$R=$Rydberg constant
${{n}_{i}}=$Initial quantum number
${{n}_{f}}=$Final quantum number
For first Balmer line we have
$\dfrac{1}{{{\lambda }_{1}}}=R\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}} \right)$ where ${{\lambda }_{1}}$ denotes the wavelength of first Balmer line.
$\Rightarrow \dfrac{1}{660}=R\left( \dfrac{1}{4}-\dfrac{1}{9} \right)$
$\Rightarrow \dfrac{1}{660}=R\left( \dfrac{9-4}{36} \right)$
$\Rightarrow\dfrac{1}{660}=R\left( \dfrac{5}{36} \right)$
$\Rightarrow\dfrac{1}{660}=\dfrac{5R}{36}$……………… (i)
Now, for second Balmer line we have
$\dfrac{1}{{{\lambda }_{2}}}=R\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{4}^{2}}} \right)$ where ${{\lambda }_{2}}$ denotes the wavelength of second Balmer line.
$\Rightarrow \dfrac{1}{{{\lambda }_{2}}}=R\left( \dfrac{1}{4}-\dfrac{1}{16} \right)$
$\Rightarrow \dfrac{1}{{{\lambda }_{2}}}=R\left( \dfrac{4-1}{16} \right)$
$\Rightarrow \dfrac{1}{{{\lambda }_{2}}}=R\left( \dfrac{3R}{16} \right)$
$\Rightarrow \dfrac{1}{{{\lambda }_{2}}}=\dfrac{3R}{16}$……………. (ii)
Now, dividing $(i)$ by $(ii)$, we get
$\Rightarrow \dfrac{\dfrac{1}{660}}{\dfrac{1}{{{\lambda }_{2}}}}=\dfrac{\dfrac{5R}{36}}{\dfrac{3R}{16}}$
$\Rightarrow \dfrac{{{\lambda }_{2}}}{660}=\dfrac{5R\times 16}{36\times 3R}$
Calculating further,
$\Rightarrow {{\lambda }_{2}}=660\times \dfrac{20}{27}$
$\Rightarrow {{\lambda }_{2}}=488.9nm$

So, the correct answer is “Option c”.

Additional Information:
Whenever an electron jumps from lower energy level to higher energy level then it gains energy and when it jumps from higher energy level to lower energy level then it loses energy. There is a relation, $\dfrac{n\left( n-1 \right)}{2}$ to find the maximum number of spectral lines when an electron jumps from ${{n}^{th}}$ orbit to ground state.

Note:
To find wavelength of hydrogen like atoms we use the following given formula:-
$\dfrac{1}{\lambda }=R{{Z}^{2}}\left( \dfrac{1}{n_{i}^{2}}-\dfrac{1}{n_{f}^{2}} \right)$ where, each symbol has the meaning as used in the above given problem. $Z$ denotes the atomic number of the hydrogen like atoms. It should also be noted that the velocity of electrons around the nucleus goes on decreasing as $n$ increases.