Question

The wavelength at which a thermal radiator at temperature $\mathit{T}$radiates electromagnetic waves most intensely is given by Wien’slaw:localid="1663130298382" ${\mathit{\lambda }}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\mathbf{(}\mathbf{2898}\mathbf{\mu }\mathbf{m}\mathbf{.}\mathbf{K}\mathbf{)}\mathbf{/}\mathit{T}$(a) Show that the energy$\mathit{E}$of a photon corresponding to that wavelength can be computed from

$\mathit{E}\mathbf{=}\mathbf{(}\mathbf{4}\mathbf{.28}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\frac{\mathbf{\text{MeV}}}{\mathbf{K}}\mathbf{)}\mathit{T}$

(b) At what minimum temperature can this photon create an electron-positron pair(as discussed in Module 21-3)?

Short answer

(a) The energy of the photon.$E=(4.28\times {10}^{-10}\text{MeV}/\text{K})T$

(b) The temperature of the photon is.$T=2.39\times {10}^9\text{K}$

Step-by-step solution

Given Information

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

${\lambda }_{\max }=\frac{(2898\text{\hspace{0.17em}μm}\cdot \text{K})}{T}$

Write the formula for the energy of the photon as:

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

Step 3: Show that the energyE of a photon corresponding to that wavelength can be computed.

(a)

Consider the given wavelength, ${\lambda }_{\max }=\frac{(2898\text{\hspace{0.17em}μm}\cdot \text{K})}{T}$and substitute it into the Equation (1) as follows:

,$\begin{array}{l}E=\frac{hc}{\lambda }\\ E=\frac{(1240\text{\hspace{0.17em}eV}\cdot \text{nm})}{\frac{(\text{2892\hspace{0.17em}μm}\cdot \text{K})}{\text{T}}}\end{array}$

Convert the units as follows:

$\begin{array}{c}{\lambda }_{\max }=\frac{(\text{2892\hspace{0.17em}μm}\cdot \text{K})}{\text{T}}\\ =2.898\times {10}^6\text{\hspace{0.17em}nmK}\end{array}$

Thus,

$\begin{array}{l}E=\frac{(1.240\times {10}^{-3}\text{\hspace{0.17em}MeV}\cdot \text{nm})T}{2.868\times {10}^6\text{\hspace{0.17em}nm}\cdot \text{K}}\\ E=(4.28\times {10}^{-10}\frac{\text{MeV}}{\text{K}})T\end{array}$

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

Determine the minimum temperature can this photon create an electron-positron pair

(b)

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

Hence for solution (a),

$\begin{array}{l}T=\frac{E}{4.28\times {10}^{-10}\frac{\text{MeV}}{\text{K}}}\\ T=\frac{1.022\text{MeV}}{4.28\times {10}^{-10}\frac{\text{MeV}}{\text{K}}}\\ T=2.39\times {10}^9\text{\hspace{0.17em}K}\end{array}$

Therefore, the minimum temperature at which the photon create an electron-positron pair is.$T=2.39\times {10}^9\text{\hspace{0.17em}K}$