Question

The ratio of difference of $1st$ and $2nd$ lines of Lyman series in H-like atom to difference in wavelength for $2nd$ and $3rd$ lines of same series is:
A. $2.5:1$
B. $3.5:1$
C. $4.5:1$
D. $5.5:1$

Answer 1

In order to solve this question we need to understand the spectrum of hydrogen. So atoms configuration as suggested by Bohr is that the electrons revolve around the nucleus in shells such that their angular momentum is integral multiple of $\dfrac{h}{{2\pi }}$ in that shell. So when electrons at higher state or shell transits to lower state then it emits light of a certain wavelength. Collection of all those wavelengths are known as spectrum. Based on order of transitions, they are named as Lyman series transitions, Balmer, Paschen, Brackett, and Pfund series transitions.

Complete step by step answer:
Let the electron transit from ${n_2}$ level to ${n_1}$ level. Let the light wavelength it emits while doing so be, $\lambda$. Let $R$ be Rydberg constant.So from definition of transition we get,
$\dfrac{1}{\lambda } = R(\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}) \to (i)$

For Lyman series transition, ${n_1} = 1$ as electrons in this case transit from higher state to first energy state.So equation (i) become,
$\dfrac{1}{\lambda } = R(1 - \dfrac{1}{{{n_2}^2}})$
So for $1st$ order transition we have, ${n_2} = 2$
Putting values we get,
$\dfrac{1}{{{\lambda _1}}} = R(1 - \dfrac{1}{{{2^2}}})$
$\Rightarrow \dfrac{1}{{{\lambda _1}}} = R\dfrac{3}{4}$
$\Rightarrow {\lambda _1} = \dfrac{4}{{3R}}$
So for $2nd$ order transition we have, ${n_2} = 3$
Putting values we get,
$\dfrac{1}{{{\lambda _2}}} = R(1 - \dfrac{1}{{{3^2}}})$
$\Rightarrow \dfrac{1}{{{\lambda _2}}} = R\dfrac{8}{9}$
$\Rightarrow {\lambda _2} = \dfrac{9}{{8R}}$

Similarly, for $3rd$ order transition we have, ${n_2} = 4$
Putting values we get,
$\dfrac{1}{{{\lambda _3}}} = R(1 - \dfrac{1}{{{4^2}}})$
$\Rightarrow \dfrac{1}{{{\lambda _3}}} = R\dfrac{{15}}{{16}}$
$\Rightarrow {\lambda _3} = \dfrac{{16}}{{15R}}$
So the difference of $1st{\kern 1pt} \& {\kern 1pt} {\kern 1pt} 2nd$ lines of Lyman series is, $\lambda = {\lambda _1} - {\lambda _2}$
Putting values we get,
$\lambda = \dfrac{4}{{3R}} - \dfrac{9}{{8R}}$
$\Rightarrow \lambda = \dfrac{{32 - 27}}{{24R}}$
$\Rightarrow \lambda = \dfrac{5}{{24R}}$

So the difference of $2nd{\kern 1pt} \& {\kern 1pt} {\kern 1pt} 3rd$ lines of Lyman series is, $\lambda ' = {\lambda _2} - {\lambda _3}$
Putting values we get,
$\lambda ' = \dfrac{9}{{8R}} - \dfrac{{16}}{{15R}}$
$\Rightarrow \lambda ' = \dfrac{{135 - 128}}{{120R}}$
$\Rightarrow \lambda ' = \dfrac{7}{{120R}}$
So the required ratio is,
$\dfrac{\lambda }{{\lambda '}} = \dfrac{{(\dfrac{5}{{24R}})}}{{(\dfrac{7}{{120R}})}}$
$\Rightarrow \dfrac{\lambda }{{\lambda '}} = \dfrac{{25}}{7}$
$\therefore \lambda :\lambda ' = 3.5:1$

The correct option is B.

Note: It should be remembered that, when the electron is at higher level it de-excites in microseconds because of time-energy Heisenberg uncertainty principle, according to which if energy is fixed or perfectly defined then uncertainty in time is maximum, this clearly depicts why electrons at higher excited level de-excites in microseconds.