Question

The $ {H_a} $ of Balmer series has a wavelength $ \lambda = $ ____nm
(A) 565 nm
(B) 595 nm
(C) 656 nm
(D) 702 nm

Answer 1

Hint
For the Balmer series, the electron travels from the orbit with the quantum number $ {n_i} = 3,4,... $ to a final orbit, $ {n_f} = 2 $ . So we can find the wavelength between these 2 orbits by the formula $ \dfrac{1}{\lambda } = {R_\alpha }{z^2}\left( {\dfrac{1}{{n_f^2}} - \dfrac{1}{{n_i^2}}} \right) $ . So, by substituting the values for the constants, initial and final orbit we can find the value of $ \dfrac{1}{\lambda } $ and taking the inverse of which we will get the wavelength.
In solving this problem, we will use the following formula,
$\Rightarrow \dfrac{1}{\lambda } = {R_\alpha }{z^2}\left( {\dfrac{1}{{n_f^2}} - \dfrac{1}{{n_i^2}}} \right) $ , this is called the Rydberg’s equation
where $ \lambda $ is the wavelength for the Balmer series, $ z $ is the atomic number of the atomic species,
 $\Rightarrow {n_i},{n_f} $ are the initial and the final orbit of the motion of the electron, $ {R_\alpha } $ is the Rydberg constant. The value of this constant is given by the formula,
 $\Rightarrow {R_\alpha } = \dfrac{{m{e^4}}}{{64{\pi ^3}\varepsilon _o^2c{\hbar ^3}}} = 1.09737 \times {10^7}{m^{ - 1}} $ .

Complete step by step answer
Whenever an electron from any species of atom jumps from one orbit to another, it absorbs or releases energy to do so. This is because various orbits are related to a specific amount of energies and the energy that the electron absorbs or emits is equal to the difference between the energies of the two orbits.
This energy is absorbed or emitted by the electron in the form of a photon. Here we are calculating the wavelength associated with that photon.
When the final orbit of an electron is 2 and the initial orbit is $ 3,4,5,... $ then the transition gives us the Balmer series. In the Balmer series, when $ {n_f} = 2 $ and $ {n_i} = 3 $ then we get the $ {H_a} $ line in the Balmer series.
Now, from the Rydberg’s equation, we have,
 $\Rightarrow \dfrac{1}{\lambda } = {R_\alpha }{z^2}\left( {\dfrac{1}{{n_f^2}} - \dfrac{1}{{n_i^2}}} \right) $
So in this case we take $ z = 1 $ and the value of the Rydberg’s constant can be calculated as,
 $\Rightarrow {R_\alpha } = \dfrac{{m{e^4}}}{{64{\pi ^3}\varepsilon _o^2c{\hbar ^3}}} $
Now we can substitute the value of all the constants where
 $\Rightarrow m = 9.1 \times {10^{ - 31}}kg $ is the mass of an electron and $ e = 1.6 \times {10^{ - 19}}C $ is the electronic charge and the values of the other constants are,
 $\Rightarrow \hbar = 1.054 \times {10^{ - 34}}N{m^2}/k{g^2} $ , $ {\varepsilon _o} = 8.854 \times {10^{ - 12}}F/m $ and $ c = 3 \times {10^8}m/s $
Substituting all these constants, we get the value of Rydberg’s constant as,
 $\Rightarrow {R_\alpha } = 1.09737 \times {10^7}{m^{ - 1}} $
So the Rydberg’s equation becomes,
 $\Rightarrow \dfrac{1}{\lambda } = 1.09737 \times {10^7}\left( {\dfrac{1}{{n_f^2}} - \dfrac{1}{{n_i^2}}} \right) $
Now for the $ {H_a} $ line, we substitute $ {n_i} = 3 $ and $ {n_f} = 2 $ .
Therefore, we get the equation as,
 $\Rightarrow \dfrac{1}{\lambda } = 1.09737 \times {10^7}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right) $
 $\Rightarrow \dfrac{1}{\lambda } = 1.09737 \times {10^7}\left( {\dfrac{1}{4} - \dfrac{1}{9}} \right) $
Now by subtracting the fractions we get,
 $\Rightarrow \dfrac{1}{\lambda } = 1.09737 \times {10^7}\left( {\dfrac{{9 - 4}}{{36}}} \right) $
 $\Rightarrow \dfrac{1}{\lambda } = 1.09737 \times {10^7} \times \dfrac{5}{{36}} $
By doing the following multiplication, we get the value of $ \dfrac{1}{\lambda } $ as,
 $\Rightarrow \dfrac{1}{\lambda } = 1524125{m^{ - 1}} $
To get $ \lambda $ we take the inverse of the calculated value, and we get
 $\Rightarrow \lambda = \dfrac{1}{{1524125}} = 6.56 \times {10^{ - 7}}m $
So this is the answer in meters. But the options are given in nanometer, so we multiply $ {10^9} $ to convert it to nanometer and we get,
 $\Rightarrow \lambda = 6.56 \times {10^{ - 7}} \times {10^9} = 656nm $ .
So the correct answer is option (C) i.e. $ \lambda = 656nm $ .

Note
We can also calculate the value of $ \lambda $ alternatively from the energy of the orbits.
For any orbit, the energy is given by,
 $\Rightarrow E = - 13.6\left( {\dfrac{{{z^2}}}{{{n^2}}}} \right)eV $
So to find the difference in energy between the two orbits, we have
 $\Rightarrow {E_i} - {E_f} = \Delta E = 13.6\left( {\dfrac{1}{{n_f^2}} - \dfrac{1}{{n_i^2}}} \right) $
By substituting the values $ {n_i} = 3 $ and $ {n_f} = 2 $ we have,
 $\Rightarrow \Delta E = 13.6\left( {\dfrac{1}{4} - \dfrac{1}{9}} \right) = 13.6 \times \dfrac{5}{{36}} = 1.89eV = 1.89 \times 1.6 \times {10^{ - 19}}V $
Now, this energy difference is equal to $ \Delta E = \dfrac{{hc}}{\lambda } $ .
 So, to find $ \lambda $ we have, $ \lambda = \dfrac{{hc}}{{\Delta E}} $ .
Substituting the values we get,
 $\Rightarrow \lambda = \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.89 \times 1.6 \times {{10}^{ - 19}}}} = 6.56 \times {10^{ - 7}}m $
 $ \therefore \lambda = 6.56 \times {10^{ - 7}} \times {10^9}nm = 656nm $ .