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.

#### Solution

Given:

Energy (E) of the ground state will be the energy acquired in the transition of the 2 excitation state to ground state.

`E_1 = (hc)/lamda_1`

Here,

*h *= Planck's constant

*c* = Speed of light

λ_{1} = Wavelength of the radiation emitted when atoms come from the highest excited state to ground state

`therefore E_1=(6.63xx10^-34)/(46xx10^-9)J`

`E_1 = ((6.63xx10^-34)xx(3xx10^8))/((46xx10^-9)xx(16xx10^-19)) eV`

= `1242/46 = 27 eV`

Energy in the first excitation state (E_2) will be the energy acquired in the transition of the highest energy state to the 2nd excitation state.

`E_2 = (hc)/lamda`

Here,

`lamda_n` = Wavelength of the radiation emitted when an atom comes from the highest energy state to the 2nd excitation state.

`E_2 = (hc)/lamda`

`E_2 = ((6.63xx10^-34)xx(3xx10^8))/(103.5xx10^-9)`

`E_2 =((6.63xx10^-34)xx(3xx10^8))/((103.5xx10^-9)xx(1.6xx10^-19)) eV`

= 12 eV