Question

Calculate energy, frequency and wavelength of the radiation which is corresponding to the spectral line of the lowest frequency in Lyman series in the hydrogen atom spectrum. Also calculate the energy for the corresponding line in the spectrum of $L{i^{2 + }}$
$\left( {{R_H} = 1.09677 \times {{10}^7}{m^{ - 1}},c = 3 \times {{10}^8}m{s^{ - 1}},Z = 3} \right)$

Answer 1

Spectral line of the lowest frequency in Lyman series in the hydrogen atom spectrum is caused due to the transition from first orbit to the second orbit. Energy, wavelength and frequency of light is interdependent.

Formulae used: $\dfrac{1}{\lambda } = {R_H}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]{Z^2}$
Where $\lambda$ is the wavelength, ${R_H}$ is the Rydberg constant, ${n_1}$ and ${n_2}$ are the lower and upper orbits involved in transition respectively and $Z$ is the atomic number of the species.
$\nu = \dfrac{c}{\lambda }$
Where $\nu$ is the frequency and $c$ is the speed of light
$E = \dfrac{{hc}}{\lambda }$
Where $E$ is the energy and $h$ is the Planck’s constant

Complete step by step answer:
To find the wavelength of the emitted light, we use the following formula:
$\dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]{Z^2}$
Where $\lambda$ is the wavelength, ${R_H}$ is the Rydberg constant, ${n_1}$ and ${n_2}$ are the lower and upper orbits involved in transition respectively and $Z$ is the atomic number of the species.
For lowest frequency in the Lyman series in the hydrogen atom spectrum, ${n_1} = 1$ and ${n_2} = 2$. Substituting the rest of the values, ${R_H} = 1.09677 \times {10^7}{m^{ - 1}}$ and $\;Z = 1$, we get:
$\dfrac{1}{\lambda } = 1.09677\left[ {\dfrac{1}{{1_{}^2}} - \dfrac{1}{{{2^2}}}} \right]{1^2}$

On simplifying, we get:
$\dfrac{1}{\lambda } = 1.097 \times {10^7}\left( {\dfrac{3}{4}} \right) = 0.82275 \times {10^7}{m^{ - 1}}$
Therefore, wavelength
$\lambda = \dfrac{1}{{0.82275 \times {{10}^7}}} = 1.216 \times {10^{ - 7}}m$
Simplifying further, we get:
$\Rightarrow \lambda = 121.6 \times {10^{ - 9}}m = 121.6nm$
As we know, $\nu = \dfrac{c}{\lambda }$
Where $\nu$ is the frequency and $c$ is the speed of light
$\Rightarrow \nu = \dfrac{{3 \times {{10}^8}m{s^{ - 1}}}}{{121.6 \times {{10}^{ - 9}}m}}$
On simplifying, we get:
$\nu = 2.46 \times {10^{15}}{s^{ - 1}}$
For finding energy, we use the formula $E = \dfrac{{hc}}{\lambda }$
Where $E$ is the energy and $h$ is the Planck’s constant
$\Rightarrow E = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{121.6 \times {{10}^{ - 9}}}}$
On solving, we get $E = 16.3 \times {10^{ - 19}}J$
As we can see from our first equation, $\dfrac{1}{\lambda } \propto {Z^2}$
$\Rightarrow \lambda \propto \dfrac{1}{{{Z^2}}}$
Therefore, to get the wavelength of the spectral line of $L{i^{2 + }}$ we just need to divide the wavelength of hydrogen atom with the square of the atomic number of Lithium (3)
${\lambda _{Li}} = \dfrac{{121.6}}{{{3^2}}}nm = 13.55nm$
Therefore, energy of this transition:
$E = \dfrac{{hc}}{{{\lambda _{Li}}}} \Rightarrow E = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{13.55 \times {{10}^{ - 9}}}}$
On solving, we get:
$E = 146.7 \times {10^{ - 19}}J$

Additional Information: Instead of wasting time calculating the wavelength of the lithium ion transition, notice how quickly we were able to solve this question as we knew about the proportionality between wavelength and atomic number of the species. Wavelengths corresponding to other series of transitions can also be found with this formula.

Note: Note that in the Lyman series, the lower orbit is always $n = 1$. Here as we want the spectral line with the lowest frequency, we want the transition with the highest wavelength (since frequency is inversely proportional to wavelength) and this is achieved in the transition to the second orbit. The transition diagram for the hydrogen atom shows all the series present with their respective transitions. The first series is the Lyman series, while the last identified series is the Humphrey-Davy series.