91Ó°ÊÓ

By Wien’s law, the wavelength at which a thermal radiator at temperature $T$radiates electromagnetic waves most intensely is.${\mathit{λ}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\mathbf{\left(}\mathbf{\text{2898μ³¾.°­}}\mathbf{\right)}\mathbf{\text{/T}}$

Consider the formulas required:

a) $\mathit{T}\mathbf{=}\frac{\mathbf{λ}}{{\mathbf{λ}}_{\mathbf{m}\mathbf{a}\mathbf{x}}}$ …… (1)

b) ${\mathit{λ}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\frac{\mathbf{λ}}{\mathbf{T}}$ ……. (2)

(a)

Consider that the cosmic background radiation peaks in intensity at a wavelength of .

For the universal microwave background, Wien’s law leads to,

$\begin{array}{l}T=\frac{2898\text{‰ӾÀ}â‹\dots \text{K}}{{\mathrm{λ}}_{\max }}\\ T=\frac{2898\text{″¾³¾}â‹\dots \text{K}}{1.1\text{″¾³¾}}\\ T=2.6\text{‿é}\end{array}$

Therefore, the given wavelength correspondsto the temperature.$T=2.6\text{‿é}$

b)

Consider that$379000\text{y}$after the big bang, the universe became transparent to electromagnetic radiation. Its temperature then was.$2970\text{K}$

At, decoupling, the wavelength can be calculate as follows:

.$\begin{array}{l}{\mathrm{λ}}_{\max }=\frac{2898\text{‰ӾÀ}â‹\dots \text{K}}{T}\\ {\mathrm{λ}}_{\max }=\frac{2898\text{‰ӾÀ}â‹\dots \text{K}}{2970\text{‿é}}\\ {\mathrm{λ}}_{\max }=0.976\text{‰ӾÀ}\end{array}$

${\mathrm{λ}}_{\max }=976\text{ n³¾}$

Therefore,the wavelength at which the background radiation was then most intenseis ${\mathrm{λ}}_{\max }=976\text{ n³¾}$.