Question

A hot gas emits radiation of wavelengths 46.0 nm, 82.8 nm and 103.5 nm only. Assume that the atoms have only two excited states and the difference between consecutive energy levels decreases as energy is increased. Taking the energy of the highest energy state to be zero, find the energies of the ground state and the first excited state.

Answer 1

Hint: When an electron jumps from higher energy level to the lower energy level then it radiates in the form of radiation. The energy of the radiation is inversely proportional to the wavelength. So, higher is the wavelength of emitted radiation lower will be the energy difference.

Formula used:
\[E = \dfrac{{hc}}{\lambda }\]
Where E is the energy of radiation having wavelength \[\lambda \] and h and c is the constant called the Plank’s constant and the speed of light respectively.

Complete step by step solution:
The wavelengths of the emitted radiation are given as 46.0 nm, 82.8 nm and 103.5 nm. As the energy difference is inversely proportional to the wavelength, so for maximum energy difference will have least wavelength. The maximum energy difference is between the ground state and the highest excited state. It is given that there are only two excited states and the energy of the second excited state is zero.

If the energy of the ground state is \[{E_0}\] then for the transition from 2nd excited state to the ground state will emit the radiation of wavelength 46.0 nm. Using energy formula,
\[{E_2} - {E_0} = \dfrac{{hc}}{{{\lambda _1}}} \\ \]
\[\Rightarrow {E_0} = - \dfrac{1}{{1.6 \times {{10}^{ - 19}}}}\left( {\dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{46 \times {{10}^{ - 9}}}}} \right)eV \\ \]
\[\Rightarrow {E_0} = - 27.0\,eV\]
Therefore, the energy of the ground state is -27.0 eV.

The energy difference between the first excited state and the second excited state will be minimum. And hence the wavelength of the radiation is maximum, i.e. 103.5nm.
Using energy formula,
\[{E_2} - {E_1} = \dfrac{{hc}}{{{\lambda _{12}}}} \\ \]
\[\Rightarrow {E_1} = - \dfrac{1}{{1.6 \times {{10}^{ - 19}}}}\left( {\dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{103.5 \times {{10}^{ - 9}}}}} \right)eV \\ \]
\[\therefore {E_1} = - 12.0\,eV\]

Therefore, the energy of the first excited state is -12.0 eV.

Note: The Rydberg formula has been used to determine the wavelengths of several spectral series that have been created from the emission spectrum of atomic hydrogen. These spectral lines that have been seen are the result of electron transitions between two energy levels within an atom. Using conservation of energy, the energy of the radiation should be equal to the energy difference between the energy levels.