Due to the presence everywhere of the cosmic background radiation, the minimum p

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 (in terms of

Explanation / Answer

a)the energy is

E = k x T

where k = 1.38 x 10^-23 J/K and T = 2.7 K

b)Also,we know that

E = (1/2)mv^2

or v^2 = (2E/m)

or v = (2E/m)^1/2

where m = 9.11 x 10^-31 kg

the wavelength of the photon is

L = h/mv

where h = 6.63 x 10^-34 J-s

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