Chapter 27, Problem 6P

(a)

To determine

The energy of the photon.

Answer to Problem 6P

The energy of the photon is $2.90\times {10}^{-19}\text{J/photon}$.

Explanation of Solution

The energy of the photon is,

$E_{Peak}=\frac{hc}{{\lambda }_{\max }}$

• • $h$ is the Plank’s constant
  • ${\lambda }_{\max }$ is the maximum wavelength
  • $c$ is the speed of the light

Substitute $6.63\times {10}^{-34}\text{J}\text{.s}$ for $h$, $3.00\times {10}^8\text{m/s}$ for $c$ and $685\text{nm}$ for ${\lambda }_{\text{max}}$.

$\begin{array}{c}{E}_{Peak}=\frac{\left(6.63\times {10}^{-34}\text{J}\text{.s}\right)\left(3.00\times {10}^8\text{m/s}\right)}{\left(685\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\\ =2.90\times {10}^{-19}\text{J/photon}\end{array}$

Conclusion:

Thus, the energy of the photon is $2.90\times {10}^{-19}\text{J/photon}$.

(b)

To determine

The surface temperature of the star.

Answer to Problem 6P

The surface temperature of the star is $4.23\times {10}^3\text{K}$.

Explanation of Solution

According to Wein’s displacement law, the temperature is,

$T=\frac{0.2898\times {10}^{-2}\text{m}\text{.K}}{{\lambda }_{\text{max}}}$

• • ${\lambda }_{\text{max}}$ is the maximum wavelength

Substitute $685\text{nm}$ for ${\lambda }_{\text{max}}$.

$\begin{array}{c}T=\frac{0.2898\times {10}^{-2}\text{m}\text{.K}}{\left(685\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\\ =4.23\times {10}^3\text{K}\end{array}$

Conclusion:

Thus, the surface temperature of the star is $4.23\times {10}^3\text{K}$.

(c)

To determine

The rate at which energy is emitted from the star in the form of radiation.

Answer to Problem 6P

The rate at which energy is emitted from the star in the form of radiation is $1.65\times {10}^{26}\text{W}$

Explanation of Solution

The power radiated by an object is,

$\begin{array}{c}P=\sigma Ae{T}^4\\ =\sigma \left(4\pi r^2\right)e{T}^4\end{array}$

• • $\sigma$ is the Stefan’s constant
  • $A$ is the surface area
  • $T$ is the temperature
  • $e$ is the emissivity
  • $r$ is the radius

Substitute $4.23\times {10}^3\text{K}$ for $T$, $5.6696\times {10}^{-8}{\text{W/m}}^2.{\text{K}}^4$ for $\sigma$, $1$ for $e$ and $8.50\times {10}^8\text{m}$ for $r$.

$\begin{array}{c}P=\left(5.6696\times {10}^{-8}{\text{W/m}}^2.{\text{K}}^4\right)4\pi {\left(8.50\times {10}^8\text{m}\right)}^2{\left(4.23\times {10}^3\text{K}\right)}^4\left(1\right)\\ =1.65\times {10}^{26}\text{W}\end{array}$

Conclusion:

Thus, the rate at which energy is emitted from the star in the form of radiation is $1.65\times {10}^{26}\text{W}$.