Question

The longest wavelength doublet absorption transition is observed at $589$and $589.6\,\,nm$. Calculate the frequency of each transition and energy difference between two excited states.

Answer 1

Hint: In question two wavelengths are given to us with the help frequency speed wavelength relation we can calculate the frequency of each transition. Using the above calculated frequency of the transition wave in the energy formula we calculate the energy of each transition then subtracting them from each other we can get the energy difference between two excited states.

Formula used:
$v = \dfrac{c}{\lambda }$
Where, Frequency of the wave $ = v$ and $\lambda$= wavelength of wave.
The frequency of the wave is ${s^{ - 1}}$.
Speed of light $ = c = 3.0 \times {10^8}m{s^{ - 1}}$
Unit of speed of light is same as that of normal speed that is $m{s^{ - 1}}$
$E = hv$
Where, energy $ = E$ and unit of energy is $J$.

Complete step by step answer:
As per the problem, a longest wavelength doublet absorption transition is observed at $589$ and $589.6\,\,nm$.Let,
${\lambda _1} = 589nm$
$\Rightarrow {\lambda _2} = 589.6nm$
We need to convert this nanometer to meter as it is its SI unit.
${\lambda _1} = 589 \times {10^{ - 9}}m$
$\Rightarrow {\lambda _2} = 589.6 \times {10^{ - 9}}m$
First we need to calculate the frequency of each transition wave we will get,
We know that,
$v = \dfrac{c}{\lambda }$
Where, Frequency of the wave $ = v$
Unit of frequency of wave is ${s^{ - 1}}$
Speed of light $ = c = 3.0 \times {10^8}m{s^{ - 1}}$
Unit of speed of light is same as that of normal speed that is $m{s^{ - 1}}$
That means the wave moves with the speed of light.
Wavelength of the wave $ = \lambda $
Unit of wavelength is $m$.

Now applying this to find the frequency of respective wavelength:
${v_1} = \dfrac{c}{{{\lambda _1}}}$
Putting the known values in the above equation we will get,
Frequency 1:
${v_1} = \dfrac{{3.0 \times {{10}^8}m{s^{ - 1}}}}{{589 \times {{10}^{ - 9}}m}}$
Further solving we will get,
${v_1} = 0.005093 \times {10^{17}}{s^{ - 1}}$
$ \Rightarrow {v_1} = 5.093 \times {10^{14}}{s^{ - 1}}$
Frequency 2:
${v_2} = \dfrac{{3.0 \times {{10}^8}m{s^{ - 1}}}}{{589.6 \times {{10}^{ - 9}}m}}$
Further solving we will get,
${v_2} = 0.005088 \times {10^{17}}{s^{ - 1}}$
$ \Rightarrow {v_2} = 5.088 \times {10^{14}}{s^{ - 1}}$
Hence,
${v_1} = 5.093 \times {10^{14}}{s^{ - 1}}$
$\Rightarrow {v_2} = 5.088 \times {10^{14}}{s^{ - 1}}$

Now using energy formula,
We know,
$E = hv$
Where, Energy $ = E$.
Unit of energy is $J$ .
Plack’s constant$ = h = 6.626 \times {10^{ - 34}}{m^2}kg\,{s^{ - 1}}$
Frequency $ = v$
Now, ${E_1} = h{v_1}$ and ${E_2} = h{v_2}$.
Subtranting energy one from every two we will get,
$\Delta E = {E_2} - {E_1}$
Where, change in energy $ = \Delta E$

Putting the respective values we will get,
$\Delta E = h{v_2} - h{v_1}$
$ \Rightarrow \Delta E = h\left( {{v_2} - {v_1}} \right)$
Putting the known values we will get,
$\Delta E = 6.626 \times {10^{ - 34}}{m^2}kg\,{s^{ - 1}}\left( {5.088 \times {{10}^{14}}{s^{ - 1}} - 5.093 \times {{10}^{14}}{s^{ - 1}}} \right)$
Further solving we get,
$\Delta E = - 6.626 \times {10^{ - 34}}{m^2}kg\,{s^{ - 1}}\left( {0.005 \times {{10}^{14}}{s^{ - 1}}} \right)$
$ \Rightarrow \Delta E = - 0.0331 \times {10^{ - 20}}{m^2}kg\,{s^{ - 2}}$
$ \Rightarrow \Delta E = - 3.31 \times {10^{ - 22}}{m^2}kg\,{s^{ - 2}}$
$ \therefore \Delta E = - 3.31 \times {10^{ - 22}}J$

Hence energy difference between two excited states $\Delta E = - 3.31 \times {10^{ - 22}}\,J$.

Note: Remember before calculating the frequency first convert the wavelength from nanometer to meter because the speed which we use while calculating the wavelength is measured in meters per second and to get a simplified frequency value. Also to get a perfect SI unit of frequency and energy. And speed of light and planck's constant will remain the same for every condition.