Question

The wavelength of the second line of the Balmer series in the hydrogen spectrum is 4816 $A^o$. The wavelength of the first line is
A) $\dfrac{{27}}{{20}} \times \;4861$ $A^o$
B) $\dfrac{{20}}{{27}} \times \;4861$ $A^o$
C) $20 \times \;4861$ $A^o$
D) $4861$ $A^o$

Answer 1

Hint: A series of spectral lines obtained when an electron makes a transition from any high energy level (n=3, 4, 5, 6….) to second energy level (p=2) is termed as the Balmer series. It lies in the visible region of the electromagnetic spectrum. The first line in the series (n=3 to p=2) is called ${{\rm{H}}_\alpha }$ line, the second line in the series (n=4 to p=2) is called ${{\rm{H}}_\beta }$ line, etc.

Formula used:
$\dfrac{1}{\lambda }\; = \,{\rm{R}}{{\rm{Z}}^2}\left( {\dfrac{1}{{{{\rm{p}}^2}}}\, - \,\dfrac{{\rm{1}}}{{{{\rm{n}}^2}}}} \right)$
Where, R is the Rydberg constant ${\rm{ = }}\;{\rm{1}}{\rm{.09737}} \times {\rm{ 1}}{{\rm{0}}^7}\;{m^{ - 1}}$
Z is the atomic number (for hydrogen Z=1)
P is the lower energy level
n is the highest energy level
Complete step by step answer:
For the first line in Balmer series
$\dfrac{1}{\lambda }\; = \;{\rm{R}}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right)\; = \;{\rm{R}}\left( {\dfrac{1}{4} - \dfrac{1}{9}} \right)\, = \;{\rm{R}}\left( {\dfrac{{9 - 4}}{{36}}} \right)\; = \;\dfrac{{{\rm{5R}}}}{{16}}$
For the second line in Balmer series
$\dfrac{1}{{4816}}\; = \;{\rm{R}}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{4^2}}}} \right)\; = \;{\rm{R}}\left( {\dfrac{1}{4} - \dfrac{1}{{16}}} \right)\, = \;{\rm{R}}\left( {\dfrac{{4 - 1}}{{16}}} \right)\; = \;\dfrac{{{\rm{3R}}}}{{16}}$
Because $\lambda $ (wavelength) is given for second line = $4816 $A^o$ $
Now divided one by second we get,
$\dfrac{\lambda }{{4816}}\; = \,\dfrac{{3{\rm{R}}}}{{16}} \times \;\dfrac{{36}}{{5{\rm{R}}}}$
 \[ \Rightarrow \lambda \; = \,4816 \times \dfrac{{20}}{{27}}\, A^o \]

Hence the correct option is (A).

Note: we should remember the definition of the Balmer series and should not get confused with orbital numbers related to the transition in the Balmer series. While considering maximum and minimum wavelengths in the Balmer series, we should estimate the final orbital with taking care of signs and relation in the formula used. Spectral lines are formed by the electronic transitions occurring between different energy levels. All the lines in Lyman series are in UV region, Balmer series are in visible region and the other four are in infrared. Electronic transition forms photons having the same energy. The energy difference between each state is fixed.