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

Explanation:

In the question, it is given that

The ground state’s energy (E) will be the energy developed in the transition to ground state of the two excitation states.

$\mathrm{E} 1=\frac{h c}{\lambda 1}$

Here,

h = Constant of Planck

c = light’s speed  

λ1 = Radiation’s wavelength emitted when atoms go to the ground state from the highest excited state,

$\text { Therefore, } E 1=(3 \times 108)\left(46 \times 10^{-9}\right) J \times\left(6.63 \times 10^{-(34)}\right)$.

$E 1=(3 \times 108)(46 \times 10-9) \times(6.63 \times 10-34) \times(1.6 \times 10-19) e V=27 e V$

In the first excitation state, the energy is E2. This is the acquired energy in the transition to the 2nd excitation state of the highest energy state.$E 2=\frac{h c}{\lambda_{n}}$

Here,  

λn = Radiation’s wavelength released when an atom comes to the 2nd excitation state from the highest energy state.

$\mathrm{E} 2=\frac{h c}{\lambda n}$

$E 2=4.14 \times 10-15 \times 3 \times \frac{108}{103.5}=12 e V$.