91Ó°ÊÓ

The wavelength at which a thermal radiator at temperature $T$radiates electromagnetic waves most intensely is given by Wien’s law as follows,

${\mathrm{λ}}_{\max }=\frac{(2898\text{‰ӾÀ}â‹\dots \text{K})}{T}$

Write the formula for the energy of the photon as:

$\mathit{E}\mathbf{=}\frac{\mathbf{h}\mathbf{c}}{\mathbf{λ}}$…… (1)

(a)

Consider the given wavelength, ${\mathrm{λ}}_{\max }=\frac{(2898\text{‰ӾÀ}â‹\dots \text{K})}{T}$and substitute it into the Equation (1) as follows:

,$\begin{array}{l}E=\frac{hc}{\mathrm{λ}}\\ E=\frac{(1240\text{ e³Õ}â‹\dots \text{nm})}{\frac{(\text{2892‰ӾÀ}â‹\dots \text{K})}{\text{T}}}\end{array}$

Convert the units as follows:

$\begin{array}{c}{\mathrm{λ}}_{\max }=\frac{(\text{2892‰ӾÀ}â‹\dots \text{K})}{\text{T}}\\ =2.898×{10}^6\text{ n³¾°­}\end{array}$

Thus,

$\begin{array}{l}E=\frac{(1.240×{10}^{−3}\text{ M±ð³Õ}â‹\dots \text{nm})T}{2.868×{10}^6\text{ n³¾}â‹\dots \text{K}}\\ E=(4.28×{10}^{−10}\text{ }\frac{\text{MeV}}{\text{K}})T\end{array}$

Therefore, It has been proved that energy of the photon corresponding to the wavelength given can be computed from .$E=(4.28×{10}^{−10}\text{MeV}/\text{K})T$

(b)

It requires twice the energy of the rest of the electrons to create an electron-positron pair.

Hence for solution (a),

$\begin{array}{l}T=\frac{E}{4.28×{10}^{−10}\text{ }\frac{\text{MeV}}{\text{K}}}\\ T=\frac{1.022\text{MeV}}{4.28×{10}^{−10}\text{ }\frac{\text{MeV}}{\text{K}}}\\ T=2.39×{10}^9\text{‿é}\end{array}$

Therefore, the minimum temperature at which the photon create an electron-positron pair is.$T=2.39×{10}^9\text{‿é}$