Question

Calculate the frequency, energy and wavelength of the radiation corresponding to the spectral line of lowest frequency in the Lyman series in the spectra of the hydrogen atom. Also, calculate the energy of the corresponding line in the spectra of $L{i^{ + 2}}$.

Answer 1

Lyman series is the one where the electron comes to shell number 1 (${n_1}$ = 1) and lowest frequency is obtained when the wavelength is the longest which means the transition should be shortest.
Formulas used:
-Wavelength:
$\dfrac{1}{\lambda } = R{z^2}\left[ {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right]$
-Frequency: $\nu = \dfrac{c}{\lambda }$
-Energy:$E = h\nu = h\dfrac{c}{\lambda }$
Where,
λ = wavelength;
R = Rydberg’s Constant = $1.0967 \times {10^7}$;
$\nu$ = frequency;
c = speed of light;
E = energy.

Complete step by step solution:
-The spectral line of lowest frequency (highest wavelength) in the Lyman series in the spectra of Hydrogen atoms has an electronic transition from shell 2 to shell 1.
This means that ${n_1}$ = 1 and ${n_2}$ = 2. Also R = $1.0967 \times {10^7}$.
-Now we need to calculate the frequency, energy and wavelength of the radiation emitted when the transition is from 2 to 1 (${n_1}$ = 1 and ${n_2}$ = 2) in an H atom.
-Let us start with this wavelength. It is calculated using the formula:
$\dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right]$
= $1.0967 \times {10^7}\left[ {\dfrac{1}{1} - \dfrac{1}{{{2^2}}}} \right]$
= $1.096 \times {10^7}\left[ {\dfrac{3}{4}} \right]$
$\lambda = \dfrac{4}{{1.096 \times {{10}^7} \times 3}}$
= $\dfrac{4}{{3.288 \times {{10}^7}}}$
= $1.2165 \times {10^{ - 7}}$ m
= 121.65 nm

So, the wavelength associated with the lowest frequency of the H atom in the Lyman series is 121.65 nm.

-We will now find out the lowest frequency using the formula:
$\nu = \dfrac{c}{\lambda }$ , where c is speed of light = $3 \times {10^8}m/\sec$
= $\dfrac{{3 \times {{10}^8}}}{{1.2165 \times {{10}^{ - 7}}}}$
= $2.466 \times {10^{15}}$ ${\sec ^{ - 1}}$
So, the lowest frequency of H atom is $2.466 \times {10^{15}}$ ${\sec ^{ - 1}}$.
-Now let us calculate the energy for this lowest frequency of H atoms. It is done using the formula:
$E = h\nu = h\dfrac{c}{\lambda }$ , where h = $6.626 \times {10^{ - 34}}$
= $(6.626 \times {10^{ - 34}})(2.466 \times {10^{15}})$
= $16.3397 \times {10^{ - 19}}$ J
So, the energy of this H atom is $16.3397 \times {10^{ - 19}}$ J.

So, the wavelength, frequency and energy associated with H atom during transition from 2 to 1 (of lowest frequency) are: 121.65 nm, $2.466 \times {10^{15}}$ ${\sec ^{ - 1}}$ and $16.3397 \times {10^{ - 19}}$ J.

-Now we will calculate the energy of the corresponding line or the radiation emitted when transition is from 2 to 1 (${n_1}$ = 1 and ${n_2}$ = 2) in $L{i^{ + 2}}$ atom.
Here z = 2 and R = $1.0967 \times {10^7}$
We will start with wavelength. The formula used for calculating wavelength is:
$\dfrac{1}{\lambda } = R{z^2}\left[ {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right]$
= $1.0967 \times {10^7} \times {(3)^2}\left[ {\dfrac{1}{1} - \dfrac{1}{{{2^2}}}} \right]$
= $1.096 \times {10^7} \times 9\left[ {\dfrac{3}{4}} \right]$
$\lambda = \dfrac{4}{{1.096 \times {{10}^7} \times 9 \times 3}}$
= $\dfrac{4}{{29.592 \times {{10}^7}}}$
= $0.135 \times {10^{ - 7}}$ m
= 13.5 nm

So, the wavelength of $L{i^{ + 2}}$ atom is 13.5 nm.

Finally we can find out the energy of $L{i^{ + 2}}$ atom. The formula used to calculate energy is:
$E = h\nu = h\dfrac{c}{\lambda }$ , where h = $6.626 \times {10^{ - 34}}$
= $6.626 \times {10^{ - 34}} \times \dfrac{{3 \times {{10}^8}}}{{0.135 \times {{10}^{ - 7}}}}$
= $147.244 \times {10^{ - 19}}$J

So, the energy of $L{i^{ + 2}}$ atom is $147.244 \times {10^{ - 19}}$ J and the energy associated with $L{i^{ + 2}}$ atom during transition from 2 to 1 is $147.244 \times {10^{ - 19}}$ J.

Note: The emission spectrum of hydrogen consists of different spectral series. The spectral lines are formed due to transition of electrons between two energy levels in an atom. The spectral series can be: Lyman (${n_1}$ = 1), Balmer (${n_1}$ = 2), Paschen (${n_1}$ = 3), Bracket (${n_1}$ = 4), P-fund (${n_1}$ = 5) and so on.