Question

he first excited state of the helium atom lies at an energy 19.82 eV above the ground state. If this excited state is three-fold degenerate while the ground state is non-degenerate, find the relative populations of the first excited and the ground states for helium gas in thermal equilibrium at 10,000 K.

Answer 1

Answer:

Relative population is  2.94 x 10⁻¹⁰.

Explanation:

Let N₁ and N₂ be the number of atoms at ground and first excited state of helium respectively and E₁ and E₂ be the ground and first excited state energy of helium respectively.

The ratio of population of atoms as a function of energy and temperature is known as Boltzmann Equation. The equation is:

$\frac{N_{1} }{N_{2} }$ =  $\frac{g_{1}e^{\frac{-E_{1} }{KT} } }{g_{2}e^{\frac{-E_{2} }{KT} }}$

$\frac{N_{1} }{N_{2} }$ = $\frac{g_{1}e^{\frac{-(E_{1}-E_{2}) }{KT} } }{g_{2}}$

Here g₁ and g₂ be the degeneracy at two levels, K is Boltzmann constant and T is equilibrium temperature.

Put 1 for g₁, 3 for g₂, -19.82 ev for (E₁ - E₂) and 8.6x10⁵ ev/K for K and 10000 k for T in the above equation.

$\frac{N_{1} }{N_{2} }$ = $\frac{1\times e^{\frac{-(-19.82)}{8.6\times 10^{-5}\times 10000} } }{3}$

$\frac{N_{1} }{N_{2} }$ = 3.4 x 10⁹

$\frac{N_{2} }{N_{1} }$ =  2.94 x 10⁻¹⁰