A hot gas emits radiation exclusively at wavelengths of 46.0 nm ,...

The energies of the emitted photons are calculated using $E = \dfrac{hc}{\lambda}$, taking $hc = 1242 \text{ eV nm}$.

$E_a = \dfrac{1242}{46.0} = 27 \text{ eV}$

$E_b = \dfrac{1242}{82.8} = 15 \text{ eV}$

$E_c = \dfrac{1242}{103.5} = 12 \text{ eV}$

The gas has three energy levels: ground state $E_1$, first excited state $E_2$, and second excited state $E_3$. The highest energy transition corresponds to $E_3 - E_1 = 27 \text{ eV}$.

The other two transitions are $E_2 - E_1$ and $E_3 - E_2$, which must be $15 \text{ eV}$ and $12 \text{ eV}$.

It is given that the energy difference between consecutive levels decreases with increasing energy, which implies $E_2 - E_1 > E_3 - E_2$.

Therefore, $E_2 - E_1 = 15 \text{ eV}$ and $E_3 - E_2 = 12 \text{ eV}$.

Taking the energy of the highest state as zero ($E_3 = 0$), we get:

$0 - E_2 = 12 \Rightarrow E_2 = -12 \text{ eV}$

$-12 - E_1 = 15 \Rightarrow E_1 = -27 \text{ eV}$

The energies of the ground state and the first excited state are $-27 \text{ eV}$ and $-12 \text{ eV}$ respectively.