Question

What hydrogen-like ion has the wavelength difference between the first lines of Balmer and Lyman series equal to $59.3nm{\text{ }}\left( {{R_H} = {\text{ }}109678cm^{- 1}} \right)$?

Answer 1

A spectral line is a dark or bright line in an otherwise uniform and continuous spectrum, resulting from emission or absorption of light in a narrow frequency range, compared with the nearby frequencies.

Complete step by step answer:
When electrons start revolving in the excited state the atom becomes unstable. To acquire stability electrons jump from the higher orbit to the lower orbit by the emission of the energy of value $hv$.
Where $\;v$ is the frequency of radiation energy or radiation photon. This radiation is emitted in the form of spectral lines.
According to the given question we have to find that in hydrogen-like ion has the wavelength difference between the first lines of Balmer and Lyman series equal to $59.3nm{\text{ }}\left( {{R_H} = {\text{ }}109678cm^{- 1}} \right)$.
So, let us solve this problem step by step:
Wavelength of 1st line in Balmer series,
$\dfrac{1}{{{\lambda _B}}} = {Z^2}{R_H}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right] = \dfrac{5}{{36}}{R_H}{Z^2}$

Or ${\lambda _B} = \dfrac{{36}}{{5{R_H}{Z^2}}}$

Wavelength of 1st line in Lyman series is,

$\dfrac{1}{{{\lambda _L}}} = {Z^2}{R_H}\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right]$

Or ${\lambda _L} = \dfrac{4}{{3 \times {R_H}{Z^2}}}$

Difference ${\lambda _B} - {\lambda _L} = 593 \times {10^{ - 7}} = \dfrac{{36}}{{5{R_H}{Z^2}}} - \dfrac{4}{{3{R_H}{Z^2}}} = \dfrac{1}{{{R_H}{Z^2}}}\left[ {\dfrac{{36}}{5} - \dfrac{4}{3}} \right]$

${Z^2} = \dfrac{{88}}{{593 \times {{10}^{ - 7}} \times 109678 \times 15}} = 9.0$

Or $Z = 3$

Hence, Hydrogen-like species is $L{i^{2 + }}$.

Note: An important point for the students to be remembered is that the Balmer Series involves transitions starting (for absorption) or ending (for emission) with the first excited state of hydrogen, while the Lyman Series involves transitions that start or end with the ground state of hydrogen; the adjacent image illustrates the atomic transitions that produce these two series in emission.