Answer:
The temperature associated with this radiation is 0.014K.
Step-by-step 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.
![\lambda_(max)=(b)/(T)](https://img.qammunity.org/2020/formulas/physics/college/f3koeljt4p9s14s0mvkp5a8qmgfdntvx3b.png)
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,
![T=(b)/(\lambda _(max) ) = (2.89 * 10^(-3) m.K )/(0.21m) =0.014K](https://img.qammunity.org/2020/formulas/chemistry/college/vb0o9zafi6g1hfdynfv7m6safawzg7pyej.png)