The surface of the Sun is at a temperature of approximately 5800 K, and radiates a peak wavelength of 500 nm.According to the Planck radiation law, what is its emitted intensity per unit wavelength at the peak? A)8.4 W/cm²∙nm B)42 W/cm²∙nm C)84 W/cm²∙nm D)8.4 x 10³W/cm²∙nm E)4.2 x 10⁷W/cm²∙nm

Question

Quizplus · Accepted Answer

The Planck radiation law is given by:$$I(\lambda, T) = \frac{2hc^2}{\lambda^5} \frac{1}{e^{hc/\lambda k_BT} - 1}$$where:* $I(\lambda, T)$ is the emitted intensity per unit wavelength at wavelength $\lambda$ and temperature $T$* $h$ is the Planck constant* $c$ is the speed of light* $k_B$ is the Boltzmann constantAt the peak wavelength, the derivative of $I(\lambda, T)$ with respect to $\lambda$ is zero. This gives:$$\frac{dI(\lambda, T)}{d\lambda} = 0 = \frac{2hc^2}{\lambda^6} \frac{5hc}{\lambda k_BT e^{hc/\lambda k_BT} - 1} - \frac{2hc^2}{\lambda^5} \frac{hc}{\lambda k_BT e^{hc/\lambda k_BT} - 1}$$Solving for $\lambda$, we get:$$\lambda_{	ext{peak}} = \frac{hc}{5k_BT}$$Substituting the given values, we get:$$\lambda_{	ext{peak}} = \frac{(6.626 	imes 10^{-34} 	ext{ J s})(3 	imes 10^8 	ext{ m/s})}{5(1.381 	imes 10^{-23} 	ext{ J/K})(5800 	ext{ K})}$$$$\lambda_{	ext{peak}} = 500 	ext{ nm}$$Now, we can calculate the emitted intensity per unit wavelength at the peak:$$I(\lambda_{	ext{peak}}, T) = \frac{2hc^2}{\lambda_{	ext{peak}}^5} \frac{1}{e^{hc/\lambda_{	ext{peak}} k_BT} - 1}$$$$I(\lambda_{	ext{peak}}, T) = \frac{2(6.626 	imes 10^{-34} 	ext{ J s})(3 	imes 10^8 	ext{ m/s})^2}{(500 	imes 10^{-9} 	ext{ m})^5} \frac{1}{e^{(6.626 	imes 10^{-34} 	ext{ J s})(3 	imes 10^8 	ext{ m/s})/(500 	imes 10^{-9} 	ext{ m})(1.381 	imes 10^{-23} 	ext{ J/K})(5800 	ext{ K})} - 1}$$$$I(\lambda_{	ext{peak}}, T) = 8.4 	ext{ W/cm}^2\cdot	ext{nm}$$Therefore, the correct answer is A.

The Surface of the Sun Is at a Temperature of Approximately