A hot gas emites radition of wavelength `46.0nm ,82.8nm and 103.5nm` only Assume that the atoms have only two excited state and the difference between consecutive energy levels decrease as energy is increased Taking the energy of the higest energy state to be zero find the energies of the ground state and the first excited state

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Energy Levels
We know that the gas emits radiation at specific wavelengths, which correspond to transitions between energy levels in an atom. We are given three wavelengths: 46.0 nm, 82.8 nm, and 103.5 nm. We assume there are only two excited states and the highest energy state (second excited state) has an energy of 0.

### Step 2: Calculate the Energy of the Ground State
The energy difference between the second excited state and the ground state can be calculated using the formula:
\[ E = -\frac{hc}{\lambda} \]
where:
- \( E \) is the energy,
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \) J·s),
- \( c \) is the speed of light (\( 3 \times 10^{8} \) m/s),
- \( \lambda \) is the wavelength in meters.

We will use the shortest wavelength (46.0 nm) to find the energy of the ground state:
\[ \lambda = 46.0 \, \text{nm} = 46.0 \times 10^{-9} \, \text{m} \]
Now, substituting the values into the energy formula:
\[ E_g = -\frac{(6.63 \times 10^{-34} \, \text{J·s})(3 \times 10^{8} \, \text{m/s})}{46.0 \times 10^{-9} \, \text{m}} \]

Calculating this gives:
\[ E_g = -\frac{1.989 \times 10^{-25}}{46.0 \times 10^{-9}} \]
\[ E_g = -4.32 \times 10^{-18} \, \text{J} \]

To convert this energy into electron volts (1 eV = \( 1.6 \times 10^{-19} \) J):
\[ E_g = \frac{-4.32 \times 10^{-18}}{1.6 \times 10^{-19}} \approx -27 \, \text{eV} \]

### Step 3: Calculate the Energy of the First Excited State
Next, we will calculate the energy of the first excited state using the longest wavelength (103.5 nm):
\[ \lambda = 103.5 \, \text{nm} = 103.5 \times 10^{-9} \, \text{m} \]
Using the same formula:
\[ E_2 = -\frac{hc}{\lambda} \]
Substituting the values:
\[ E_2 = -\frac{(6.63 \times 10^{-34})(3 \times 10^{8})}{103.5 \times 10^{-9}} \]

Calculating this gives:
\[ E_2 = -\frac{1.989 \times 10^{-25}}{103.5 \times 10^{-9}} \]
\[ E_2 = -1.92 \times 10^{-18} \, \text{J} \]

Converting this to electron volts:
\[ E_2 = \frac{-1.92 \times 10^{-18}}{1.6 \times 10^{-19}} \approx -12 \, \text{eV} \]

### Final Energies
- Ground State Energy \( E_g \approx -27 \, \text{eV} \)
- First Excited State Energy \( E_2 \approx -12 \, \text{eV} \)

### Summary
- The energy of the ground state is approximately \(-27 \, \text{eV}\).
- The energy of the first excited state is approximately \(-12 \, \text{eV}\).