Answer:
The temperature associated with this radiation is 0.014K.
Explanation:
If we assume that the astronomical object behaves as a black body, the relation between its wavelength and temperature is given by Wien's displacement law.
[tex]\lambda_{max}=\frac{b}{T}[/tex]
where,
λmax is the wavelength at the peak of emission
b is Wien's displacement constant (2.89×10⁻³ m⋅K)
T is the absolute temperature
For a wavelength of 21 cm,
[tex]T=\frac{b}{\lambda _{max} } = \frac{2.89 \times 10^{-3} m.K }{0.21m} =0.014K[/tex]
