Question

For Balmer series, wavelength of first line is $'{\lambda _1}'$ and for Brackett series, wavelength of first line is $'{\lambda _2}'$then $\dfrac{{{\lambda _1}}}{{{\lambda _2}}}$ is
A) 0.081
B) 0.162
C) 0.198
D) 0.238

Answer 1

Hint
The wavelength for the transition from one orbit to other is given by $\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$. First find the wavelength for the first line of the Balmer series from $n = 3$ to $n = 2$. Similarly the wavelength of the first line of Brackett series from $n = 5$ to $n = 4$ is calculated. Now divide the first wavelength by the second to get the ratio.

Complete step by step answer
When the electron is transition from one energy level to another that is to is given by,

$\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
For first line of Balmer series from $n = 3$ to $n = 2$,
$\dfrac{1}{{{\lambda _1}}} = R\left( {\dfrac{1}{{2_{}^2}} - \dfrac{1}{{3_{}^2}}} \right)$
${\lambda _1} = \dfrac{{7.2}}{R}$
For first line of Brackett series,
$\dfrac{1}{{{\lambda _2}}} = R\left( {\dfrac{1}{{{4^2}}} - \dfrac{1}{{{5^2}}}} \right)$
${\lambda _2} = \dfrac{{44.44}}{R}$
The ratio of $\dfrac{{{\lambda _1}}}{{{\lambda _2}}}$ ,
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{7.2}}{R} \times \dfrac{R}{{44.44}}$
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = 0.162$
Hence, the ratio equal to 0.162 and the correct option is (B)

Note
When an electron makes transition from higher energy level to a lower energy level then a photon of frequency $\upsilon $is emitted.
Energy of the emitted radiation is given by,
$\Delta E = 13.6{Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
Frequency of the emitted radiation is given by,
$\upsilon = Rc{Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
Recoiling of an atom is given by,
$\operatorname{Re} coil = hR{Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$