Question

A $12.3eV$ electron beam is used to bombard gaseous hydrogen at room temperature. Up to which energy level the hydrogen atoms would be excited? Calculate the wavelength of the second member of the Lyman series and the second member of Balmer series.

Answer 1

Hint Using the formula for the energy of an atom in the $n^th$ state, we need to find the energy that is required to excite the atom from the initial to the final state. This energy will be provided by the electron beam. Hence from there we can find the final state. Then using Rydberg’s formula for hydrogen we can find the wavelength of the second member of the Lyman and Balmer series.

Formula Used: In this solution we will be using the following formula,
$\Rightarrow E = \dfrac{{ - 13.6{z^2}}}{{{n^2}}}eV$
where $E$ is the energy of the atom in the $n$ th state and $z$ is the atomic number.
$\Rightarrow \dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{n_2^2}} - \dfrac{1}{{n_1^2}}} \right]$
where $\lambda$ is the wavelength, $R$ is the Rydberg constant and ${n_1}$ is the initial and ${n_2}$ is the final state.

Complete step by step answer
The energy of an atom in $n$ th state is given by the formula,
$\Rightarrow E = \dfrac{{ - 13.6{z^2}}}{{{n^2}}}eV$
Now here in the question we are considering the case for hydrogen so $z$ has a value of 1. So we get the energy as,
$\Rightarrow E = \dfrac{{ - 13.6}}{{{n^2}}}eV$
Let the initial state be ${n_i}$ and the final state be ${n_f}$ . So the energy of the initial and the final state are given as,
$\Rightarrow {E_i} = \dfrac{{ - 13.6}}{{{n_i}^2}}eV$ and ${E_f} = \dfrac{{ - 13.6}}{{{n_f}^2}}eV$
Now the energy difference between these two states is given as,
$\Rightarrow \Delta E = {E_f} - {E_i}$
So substituting these values we get,
$\Rightarrow \Delta E = \left( {\dfrac{{ - 13.6}}{{{n_f}^2}} - \dfrac{{ - 13.6}}{{{n_i}^2}}} \right)eV$
Now this difference in energy is provided by the electron beam. So we have,
$\Rightarrow 12.3 = \left( {\dfrac{{ - 13.6}}{{{n_f}^2}} - \dfrac{{ - 13.6}}{{{n_i}^2}}} \right)eV$
Initially the electron is in the ground state, so ${n_i}$ is 1. So we can write,
$\Rightarrow 12.3 = - 13.6\left( {\dfrac{1}{{{n_f}^2}} - \dfrac{1}{1}} \right)eV$
Therefore, on calculating we get,
$\Rightarrow \dfrac{1}{{{n_f}^2}} - 1 = - \dfrac{{12.3}}{{13.6}}$
Now we can take the 1 to the RHS and get,
$\Rightarrow \dfrac{1}{{{n_f}^2}} = - \dfrac{{12.3}}{{13.6}} + 1$
On the RHS by adding we get,
$\Rightarrow \dfrac{1}{{{n_f}^2}} = \dfrac{{ - 12.3 + 13.6}}{{13.6}}$
On taking inverse, we have
$\Rightarrow {n_f}^2 = \dfrac{{13.6}}{{1.3}}$
On doing the division we get
$\Rightarrow {n_f}^2 = 10.46$
On taking root,
$\Rightarrow {n_f} = \sqrt {10.46} = 3.2$
Since the state cannot be a fraction, so the hydrogen can be excited upto the third state.
Now for the second part we can use the Rydberg’s formula we have,
$\Rightarrow \dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{n_2^2}} - \dfrac{1}{{n_1^2}}} \right]$
For the Lyman series, the final state of the electron is always 1 and the initial state can be 2, 3, 4,….
For the second member, the initial state is 3. So we have ${n_1} = 3$ and ${n_2} = 1$ . The value of the Rydberg constant is $R = 1.09737 \times {10^7}{m^{ - 1}}$
So substituting the values we get,
$\Rightarrow \dfrac{1}{\lambda } = 1.09737 \times {10^7}\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{3^2}}}} \right]$
On taking LCM as 9
$\Rightarrow \dfrac{1}{\lambda } = 1.09737 \times {10^7}\left[ {\dfrac{{9 - 1}}{9}} \right]$
On doing the product we get
$\Rightarrow \dfrac{1}{\lambda } = 0.97544 \times {10^7}$
On taking inverse,
$\Rightarrow \lambda = 1.025 \times {10^{ - 7}}m$
This is the wavelength of the second member of the Lyman series.
For the Balmer series, for the second member, the initial state is ${n_1} = 4$ and the final state is ${n_2} = 2$
Now substituting the values again,
$\Rightarrow \dfrac{1}{\lambda } = 1.09737 \times {10^7}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{4^2}}}} \right]$
On taking LCM as 16
$\Rightarrow \dfrac{1}{\lambda } = 1.09737 \times {10^7}\left[ {\dfrac{{4 - 1}}{{16}}} \right]$
On doing the product we get
$\Rightarrow \dfrac{1}{\lambda } = 0.2057 \times {10^7}$
On taking inverse,
$\Rightarrow \lambda = 4.86 \times {10^{ - 7}}m$
This is the wavelength of the second member of the Balmer series.

Note
In atomic physics, Rydberg's formula is used to calculate the value of the wavelength of various chemical species. It is generally given by, $\dfrac{1}{\lambda } = R{z^2}\left[ {\dfrac{1}{{n_2^2}} - \dfrac{1}{{n_1^2}}} \right]$ . Here we have substituted the value of the atomic number as 1 as we are working with hydrogen.