# 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 - Physics

Sum

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,

= Planck's constant

c = Speed of light

λ1 = 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

Concept: Introduction of Atoms
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 21 Bohr’s Model and Physics of Atom
Q 16 | Page 384