Question

Question: Due to the presence everywhere of the cosmic background radiation, the minimum possible temperature of a gas in interstellar or intergalactic space is not 0 K but 2.7 K. This implies that a significant fraction of the molecules in space that can be in a low-level excited state may, in fact, be so. Subsequent de-excitation would lead to the emission of radiation that could be detected. Consider a (hypothetical) molecule with just one possible excited state. (a) What would the excitation energy have to be for 25% of the molecules to be in the excited state? (Hint: See Eq. 40-29.) (b) What would be the wavelength of the photon emitted in a transition back to the ground state?

Short answer

a )the excitation energy=0.1862mev

b)wavelength of the photon=6066mm

Step-by-step solution

Given information 

temperature of a gas in interstellar(T) =2.7k

hc=1240 eV

k=8.62×10^-5

concept of interstellar space and excitation space  

Interstellar space is often called the space between the stars, but more specifically, it’s the region between our Sun’s heliosphere and the astrospheres of other stars. Excitation is the discrete amount of energy (called excitation energy) to a system—such as an atomic nucleus, an atom, or a molecule—those results in its alteration, ordinarily from the condition of lowest energy (ground state) to one of higher energy (excited state).

calculation of excitation energy and wavelength of the photon emitted 

(a)

$\begin{array}{l}=18.62\times 10-5_{e}v\\ =01862m\mathrm{m}\end{array}$

(b)

$$\begin{gathered} hc=1240eVnm \\ \hspace{0.17em}\hspace{0.17em}\lambda =\frac{hc}{∆E} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1240}{18.62\times {10}^{-5}} \\ =6.66mm \end{gathered}$$