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 31% of the molecules to be in the excited state? (b) What would be the wavelength of the photon emitted in a transition back to the ground state?

Answer 1

We know that, E = K.E

E = (3/2) k T

where, k = boltzmann's constant = 1.38 x 10^-23 J/K

T = temperature of a gas in interstellar space = 2.7 K

then, we get

E = [(1.5) (1.38 x 10^-23 J/K) (2.7 K)]

E = 5.59 x 10^-23 J

(a) What would the excitation energy have to be for 31% of the molecules to be in the excited state?

E_excited = (0.31) E

E_excited = [(0.31) (5.59 x 10^-23 J)]

E_excited = 1.73 x 10^-23 J

(b) What would be the wavelength of the photon emitted in a transition back to the ground state?

using a formula, we have

E = h f = h c /

where, c = speed of light = 3 x 10⁸ m/s

h = planck's constant = 6.62 x 10^-34 J.s

then, we get

= [(6.62 x 10^-34 J.s) (3 x 10⁸ m/s)] / (1.73 x 10^-23 J)

= [(1.986 x 10^-25 J.m) / (1.73 x 10^-23 J)]

= 0.0114 m