Question

The first excited electronic energy level of the helium atom is 3.13 x 10-18 J above the ground level. Estimate the temperature at which the electronic motion will begin to make a significant contribution to the heat capacity. That is, at what temperature will the ratio of the population of the first excited state to the ground state be 5.0%?

Answer 1

Answer:

75603.86473 K

Explanation:

Given that:

The 1st excited electronic energy level of He atom = 3.13 × 10⁻¹⁸  J

The objective of this question is to estimate the temperature at which the ratio of the population will be 5.0 between the first excited state to the ground state.

The formula for estimating the ratio of population in 1st excited state to the ground state can be computed as:

$\dfrac{N_2}{N_1} = e ^{^{-\dfrac{(E_2-E_1)}{KT}}} = e ^{^{-\dfrac{(\Delta E)}{KT}}}$

From the above equation:

Δ E = energy difference =  3.13 × 10⁻¹⁸  J

k = Boltzmann constant = 1.38 × 10⁻²³  J/K

$\dfrac{N_2}{N_1} = 0.5$

Thus:

$0.05 =e^{^{ -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}$

$In (0.05) = { -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}$

$-3.00 = { -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}$

$-3.00 = -226811.5942 \times \dfrac{1}{T}$

$T = \dfrac{-226811.5942}{-3.00 }$

T = 75603.86473 K