Question

If ${\lambda _1}$and ${\lambda _2}$are the wavelengths of the first members of the Lyman and Paschen series respectively then ${\lambda _1}:{\lambda _2}$is:
A. 1 : 3
B. 1 : 30
C. 7 : 50
D. 7 : 108

Answer 1

The wavelength of light emitted in case of Hydrogen Atom when an electron moves from a higher orbit to lower orbit is given as follows:
$= > \dfrac{1}{{{\lambda _{}}}} = R(\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}})$
Where,
${\lambda _{}}$is the wavelength of emitted light
n1 and n2 are the stable energy levels in which the electrons can revolve.
The Lyman series produces spectrum in the ultraviolet region whereas the Paschen series produces spectrum in the infrared region.

Complete step by step solution:
The values of n1 and n2 for the first member of Lyman series is given as,
${n_1} = 1$
${n_2} = 2$
So, the value of ${\lambda _1}$for first member of Lyman series is given as,
$= > \dfrac{1}{{{\lambda _1}}} = R(\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}})$
Inserting the values of n1 and n2 in the above equation,
We get,
$= > \dfrac{1}{{{\lambda _1}}} = R(\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}})$
$= > \dfrac{1}{{{\lambda _1}}} = R(\dfrac{1}{1} - \dfrac{1}{4})$
$= > \dfrac{1}{{{\lambda _1}}} = R\dfrac{3}{4}$
$= > {\lambda _1} = \dfrac{4}{{3R}}$ Equation 1

The values of n1 and n2 for the first member of Paschen series is given as,
${n_1} = 3$
${n_2} = 4$
So, the value of ${\lambda _2}$for first member of Lyman series is given as,
$= > \dfrac{1}{{{\lambda _2}}} = R(\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}})$
Inserting the values of n1 and n2 in the above equation,
We get,
$= > \dfrac{1}{{{\lambda _2}}} = R(\dfrac{1}{{{3^2}}} - \dfrac{1}{{{4^2}}})$
$= > \dfrac{1}{{{\lambda _2}}} = R(\dfrac{1}{9} - \dfrac{1}{{16}})$
$= > \dfrac{1}{{{\lambda _2}}} = R\dfrac{7}{{144}}$
$= > {\lambda _2} = \dfrac{{144}}{{7R}}$ Equation 1
Finally, we need to compute the ratio ${\lambda _1}:{\lambda _2}$
So,
Dividing equation 1 by equation 2,
We get,
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{\dfrac{4}{{3R}}}}{{\dfrac{{144}}{{7R}}}}$
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{4}{{3R}} \times \dfrac{{7R}}{{144}}$
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{7}{{108}}$

Hence Option (D) is correct.

Note: One must be able to recall the values of n1 and n2 for Laymen and Paschen series in order to solve this question. Care must be taken that there is a high chance of committing silly mistakes in the calculations (since this question was calculation-intensive).