Question

What is the difference in energy between the lowest energy state of the hydrogen atom, the next higher energy level (i.e. the difference in energy between the $ 2 $ lowest energy states)?

Answer 1

Hint: To solve this question, we will use the Rydberg equation or formula. This formula is used for calculating the wavelength of the spectral line of chemical elements. The formula was initially presented as a generalization of the Balmer series for all atomic electron transitions of hydrogen.
 $ \dfrac{1}{\lambda } = {R_H}(\dfrac{1}{{{n_f}^2}} - \dfrac{1}{{{n_i}^2}}) $
Where, $ \lambda $ is the wavelength in $ m $ , which is corresponding to the energy difference between quantum levels $ {n_i} $ and $ {n_f} $ .
 $ n $ represents the principal quantum number as usual.
 $ {R_H} $ is the Rydberg constant, whose value is $ 10973731.6\;{m^{ - 1}} $ .
For solving the difference in energy we also need another formula, i.e.
 $ \Delta E = h\nu $
 $ \Rightarrow \Delta E = \dfrac{{hc}}{\lambda } $
Where, $ \Delta E $ is the difference in the energy,
 $ h $ represents Planck’s constant, which is $ 6.626 \times {10^{ - 34}}J $ ,
 $ \nu $ is the frequency which is correspondent to the energy difference,
 $ c $ is the speed of the light whose value is $ 2.998 \times {10^8}m{s^{ - 1}} $ .

Complete answer:
For solving this question, we have to follow the given three step which are:
Step-1: Firstly we have to calculate the wavelength by using Rydberg formula, i.e.
 $ \dfrac{1}{\lambda } = {R_H}(\dfrac{1}{{{n_f}^2}} - \dfrac{1}{{{n_i}^2}}) $
 $ \Rightarrow \dfrac{1}{\lambda } = \left( {10973731.6{m^{ - 1}}} \right) \times \left| {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{1^2}}}} \right| $
 $ \Rightarrow \dfrac{1}{\lambda } = 8230298.7{m^{ - 1}} $
Step-2: Now we will substitute the value in the Planck-Einstein relation, then:
 $ \Delta E = \dfrac{{hc}}{\lambda } $
 $ \Rightarrow \Delta E = hc \times \left( {8230298.7{m^{ - 1}}} \right) $
 $ \Rightarrow \Delta E = \left( {6.626 \times {{10}^{ - 34}}J} \right)\left( {2.998 \times {{10}^8}m{s^{ - 1}}} \right)\left( {8230298.7{m^{ - 1}}} \right) $
 $ \Rightarrow \Delta E = 1.635 \times {10^{ - 18}}J $
Step-3: Now we will convert the given value in $ eV $ , as we know that:
 $ 1J = \dfrac{{1eV}}{{1.602 \times {{10}^{ - 19}}J}} $
Hence by using this conversion, the value of $ \Delta E $ will be:
  $ \Delta E = \left( {1.635 \times {{10}^{ - 18}}J} \right) \times \dfrac{{1eV}}{{1.602 \times {{10}^{ - 19}}J}} $
 $ \Rightarrow \Delta E \approx 10.2eV $
This is the required answer.

Note:
For rechecking your answer you can use the energy level of hydrogen at $ n = 1 $ and $ n = 2 $ . The lowest energy level of a hydrogen atom, i.e. $ n = 1 $ is $ - 13.6eV $ and at $ n = 2 $ the energy level is $ - 3.4eV $ . Hence the difference in energy level will be $ 10.2eV $ .