Question

The Sun is approximately an ideal blackbody radiator with surface temperature of 5800 K.

(a) Find the wavelength at which its spectral radiancy is maximum and

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

(c) As we shall discuss in chapter 44, the universe is approximately an ideal blackbody radiator with radiation emitted when atoms first formed. Today the spectral radiancy of that radiation peaks at a wavelength of 1.06 mm (in the microwave region). What is the corresponding temperature of the universe?

Short answer

Thus, (a) the wavelength of spectral radiancy is maximum at $0.0388\text{\hspace{0.17em}nm}$.

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

(c) the wavelength of neutron is $9.06\times {10}^{-13}m$.

Step-by-step solution

(a) The wavelength of spectral radiancy is maximum.

Use the value $hc=1240\text{eV}\cdot \text{nm}$formula then it gives;

$\begin{array}{c}\lambda =\frac{h}{p}\\ =\frac{h}{\sqrt{2m_{e}K}}\\ =\frac{hc}{\sqrt{2m_e{c}^{2}K}}\\ =\frac{1240}{\sqrt{2\left(511000\right)\left(1000\right)}}\\ =0.0388\text{\hspace{0.17em}nm}\end{array}$

Hence, the wavelength of spectral radiancy is maximum at $0.0388\text{\hspace{0.17em}nm}$.

(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 corresponding temperature of universe.

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

$\begin{array}{c}\lambda =\frac{h}{\sqrt{2m_{n}K}}\\ =-\frac{(6.63\times {10}^{-34}J\cdot s)}{\sqrt{2(1.675\times {10}^{-27}kg)(1.6\times {10}^{-16}J)}}\\ =9.06\times {10}^{-13}m\end{array}$

Hence, the wavelength of neutron is $9.06\times {10}^{-13}m$.