Question

Just after detonation, the fireball in a nuclear blast is approximately an ideal blackbody radiator with a surface temperature of about $\mathbf{1}\mathbf{.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathit{K}$.

(a) Find the wavelength at which the thermal radiation is maximum and (b) identify the type of electromagnetic wave corresponding to that wavelength. This radiation is almost immediately absorbed by the surrounding air molecules, which produces another ideal blackbody radiator with a surface temperature of about $\mathbf{1}\mathbf{.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathit{K}$.

(c) Find the wavelength at which the thermal radiation is maximum and (d) identify the type of electromagnetic wave corresponding to that wavelength.

Short answer

Thus, (a) the wavelength of thermal radiation is maximum at$7.75\text{\hspace{0.17em}pm}$ .

(b) the wavelength is $1.24\text{\hspace{0.17em}nm}$.

(c) the wavelength of thermal radiation is $7.75\text{\hspace{0.17em}pm}$.

(d) the wavelength is $1.24\text{\hspace{0.17em}nm}$.

Step-by-step solution

(a) The wavelength of thermal radiation is maximum.

Use the value formula then it gives by the de Broglie wavelength of the electron is:

$\begin{array}{c}\lambda =\frac{h}{p}\\ =\frac{h}{\sqrt{2m_{e}K}}\\ =\frac{hc}{\sqrt{2m_{e}eV}}\\ =\frac{6.626\times {10}^{-34}Js}{\sqrt{2(9.109\times {10}^{-31}kg)(1.602\times {10}^{-19}C)(2\dot{5}.0\times {10}^{3}V)}}\\ =7.75\times {10}^{-12}m\\ =7.75\text{\hspace{0.17em}pm}\end{array}$

Hence, the wavelength of thermal radiation is maximum at $7.75\text{\hspace{0.17em}pm}$.

(b) The type of magnetic wave corresponding to the wavelength.

A photon’s de Broglie wavelength is equal to its familiar wave relationship value. Use the value$hc=1240\text{eV}\cdot \text{nm}$ gives:

$\begin{array}{c}\lambda =\frac{hc}{E}\\ =\frac{1240eV\cdot nm}{1.00keV}\\ =1.24\text{\hspace{0.17em}nm}\end{array}$

Hence, the wavelength is $1.24\text{\hspace{0.17em}nm}$.

(c) The wavelength of thermal radiation is maximum at temperature 1.0×105K. 

The neutron mass equals to $1.675\times {10}^{-27}kg$. Using the conversion from electronvolts to Joules gives;

$\begin{array}{c}\lambda =\frac{h}{p}\\ =\frac{h}{\sqrt{2m_{e}K}}\\ =\frac{hc}{\sqrt{2m_{e}eV}}\\ =\frac{6.626\times {10}^{-34}Js}{\sqrt{2(9.109\times {10}^{-31}kg)(1.602\times {10}^{-19}C)(2\dot{5}.0\times {10}^{3}V)}}\\ =7.75\times {10}^{-12}m\\ =7.75\text{\hspace{0.17em}pm}\end{array}$

Hence, the wavelength of thermal radiation is$7.75\text{\hspace{0.17em}pm}$ .

(d) The type of magnetic wave corresponding to the wavelength at temperature 1.0×105K

A photon’s de Broglie wavelength is equal to its familiar wave relationship value. Use the value$hc=1240\text{eV}\cdot \text{nm}$ gives:

$\begin{array}{c}\lambda =\frac{hc}{E}\\ =\frac{1240eV\cdot nm}{1.00keV}\\ =1.24\text{\hspace{0.17em}nm}\end{array}$

Hence, the wavelength is $1.24\text{\hspace{0.17em}nm}$.