Question

How do you calculate ionization energy using Rydberg constant?

Answer 1

Hint: In this question we have to calculate the wavelength of the absorbed photon by using Rydberg constant. After that we will substitute the value of $ \lambda $ in the Planck energy formula, hence by solving it, we get the value of $ E $ .
The Rydberg equation for absorption is:
 $ \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{n_i^2}} - \dfrac{1}{{n_f^2}}} \right) $
Where, $ \lambda $ is the wavelength of the absorbed photon, $ R $ is the Rydberg constant, $ {n_i} $ is the energy level of electron at initial and $ {n_f} $ is the energy level of electron at final.
And,
 $ E = \dfrac{{hc}}{\lambda } $
Where, $ E $ is the ionization energy, $ h $ is the Planck constant and $ c $ is the speed of the light, i.e. $ 3 \times {10^8}m{\sec ^{ - 1}} $ .

Complete Step By Step Answer:
By using the Rydberg formula, we can calculate the wavelengths of the spectral line in many elements. The formula was mostly presented as the generalization of the Balmer series for all atomic electron transitions of hydrogen. It was first stated by Johannes Rydberg in $ 1888 $ .
Hence by using the formula of Rydberg which is:
 $ \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{n_i^2}} - \dfrac{1}{{n_f^2}}} \right) $
As we know that the electron goes to infinity with respect to the atom, i.e. it leaves the atom while releasing the ionization energy. So, we have $ {n_f} = \infty $ . Assuming that we ionize from the ground state that we have $ {n_i} = 1 $ . So, after substituting these in Rydberg formula we get:
 $ \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{n_i^2}} - \dfrac{1}{{n_f^2}}} \right) $
 $ \Rightarrow \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{\infty }} \right) $
 $ \Rightarrow \dfrac{1}{\lambda } = R\left( {1 - 0} \right) $
 $ \Rightarrow \dfrac{1}{\lambda } = R $
By substituting this value in Planck energy formula we get:
 $ E = \dfrac{{hc}}{\lambda } $
 $ \Rightarrow E = hcR $
 $ \Rightarrow E = 6.626 \times {10^{ - 34}} \times 3 \times {10^8} \times 1.097 \times {10^7} $
 $ \Rightarrow 2.182 \times {10^{ - 18}}J $
Here we are dealing with electrons so the energy should be in $ eV $ . Thus,
 $ 1eV = 1.6 \times {10^{ - 19}}J $
Hence, by using this we will convent the value in $ eV $ :
 $ E = 2.182 \times {10^{ - 18}}J $
 $ \Rightarrow E = \dfrac{{2.182 \times {{10}^{ - 18}}}}{{1.6 \times {{10}^{ - 19}}}} $
 $ \Rightarrow E = 13.6eV $
Thus, the ionization energy is $ 13.6eV $ .

Note:
The Rydberg formula generally provides incorrect results for the other spectral transitions in multi-electron atoms. Since the magnitude of the screening of the inner electrons for outer electrons transitions is variable.