Question

Calculate the ionisation energy of the hydrogen atom. How much energy will be required to ionise 1 mole of hydrogen atoms? Given, the Rydberg constant is $1.0974 \times {10^7}{m^{ - 1}}$.

Answer 1

Ionisation energy refers to the amount of energy required to remove electrons from the valence shell of an atom.

Complete step by step answer:
We know that ionization energy of an atom refers to the amount of energy required to remove electrons from the valence shell of an atom. Now according to the Bohr model, we know that.

$\dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right]$………………………….(1)
Here $\lambda$ is the wavelength, R is the Rydberg constant, ${n_1}$ represents that principal quantum number of the orbital that is lower in energy and ${n_2}$ represents the principal quantum number of the orbital that is higher in energy. In the question, we were talking about ionization energy of $H - atom$, thus value of ${n_1}$ will be an and ${n_2}$ will be $\infty$ because in ionisation, electron is withdraw from ground state valence shall to the infinites value. The value of R given is $1.0974 \times {10^7}{m^{ - 1}}$. Putting all the value in eq. (1). We get-
$\dfrac{1}{\lambda } = 1.0974 \times {10^7}{m^{ - 1}}\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{\infty ^2}}}} \right]$
$or = \;\lambda \Rightarrow \dfrac{1}{{1.0974 \times {{10}^7}}}\left[ 1 \right] = 9.116 \times {10^{ - 8}}m$
$\left[ {value\;{\text{of }}\dfrac{1}{{{\infty ^2}}} = 0} \right]$
Now according to 3rd postulate of Bohr model we know that
$E = \dfrac{{hc}}{\lambda }$
Here h is the planck constant whose value is $6.63 \times {10^{ - 34}}JS$, C is the opened of light whose value is $3 \times {10^8}m{s^{ - 1}}$ and $\lambda$ is the wavelength that we have already evaluated as $9.116 \times {10^{ - 8}}$m. putting all values in above equation, we got the Energy ‘E’ of ionization energy for 1 hydrogen atom i.e.

$E = \dfrac{{6.63 \times {{10}^{ - 34}}JS \times 3 \times {{10}^8}m{s^{ - 1}}}}{{9.116 \times {{10}^{ - 8}}m}} = 2.181 \times {10^{ - 18}}J$
Which is the ionization energy for 1-hydrogen atom. Now the energy required to ionise 1 mole of hydrogen atom is multiply energy to ionise a hydrogen atom with $6.022 \times {10^{23}}$photons,
Energy ionise Required to
Ionise 1 mole of hydrogen=$2.18 \times {10^{ - 18}}J \times 6.022 \times {10^{23}}$
$= 1313 \times {10^3}mo{l^{ - 1}}$
=$1313KJmo{l^{ - 1}}$
Thus we can say that for 1 mole of hydrogen atom, energy required to ionise it is $1313KJmo{l^{ - 1}}$.
Note:
Since in this question, the lower energy state is 1, so it belongs to the lyman series question. Its lower energy state is 2,3,4 then corresponding series , where blamer, paschen, brackets series respectively.