Question

At a particular temperature T, Planck's energy density of black body radiation in terms of frequency is \(\rho_T(\nu) = 8 \times 10^{-18} \text{ J/m}^3 \text{ Hz}^{-1}\) at \(\nu = 3 \times 10^{14}\) Hz. Then Planck's energy density \(\rho_T(\lambda)\) at the corresponding wavelength (\(\lambda\)) has the value \rule{1cm}{0.15mm} \(\times 10^2 \text{ J/m}^4\). (in integer) 
[Speed of light \(c = 3 \times 10^8\) m/s] 
(Note: The unit for \(\rho_T(\nu)\) in the original problem was given as J/m³, which is dimensionally incorrect for a spectral density. The correct unit J/(m³·Hz) or J·s/m³ is used here for the solution.)

Hint:

When converting between spectral densities like \(\rho(\nu)\) and \(\rho(\lambda)\), remember the key relation \(\rho(\lambda) |d\lambda| = \rho(\nu) |d\nu|\). This leads to \(\rho(\lambda) = \rho(\nu) |\frac{d\nu}{d\lambda}|\). A common mistake is to simply substitute \(\nu=c/\lambda\) into the function, which is incorrect because it ignores the change in the differential element (\(d\nu\) vs \(d\lambda\)).

Answer 1

Correct Answer: 24

Step 1: Understanding the Concept: 
The energy distribution of blackbody radiation can be described as a function of either frequency (\(\nu\)) or wavelength (\(\lambda\)). The spectral energy density in terms of frequency, \(\rho_T(\nu)\), gives the energy per unit volume per unit frequency interval. The spectral energy density in terms of wavelength, \(\rho_T(\lambda)\), gives the energy per unit volume per unit wavelength interval. There is a direct mathematical relationship between these two quantities. 
Step 2: Key Formula or Approach: 
The energy contained in a small interval must be the same whether described by frequency or wavelength. Thus, the magnitude of the energy densities are related by: \[ \rho_T(\lambda) |d\lambda| = \rho_T(\nu) |d\nu| \] This gives the conversion formula: \[ \rho_T(\lambda) = \rho_T(\nu) \left| \frac{d\nu}{d\lambda} \right| \] Since \(\nu = c/\lambda\), we have \(\frac{d\nu}{d\lambda} = -\frac{c}{\lambda^2}\). So, \(\left| \frac{d\nu}{d\lambda} \right| = \frac{c}{\lambda^2} = \frac{\nu^2}{c}\). The conversion formula becomes: \[ \rho_T(\lambda) = \rho_T(\nu) \frac{\nu^2}{c} \] Step 3: Detailed Explanation: 
We are given: Spectral energy density in frequency, \(\rho_T(\nu) = 8 \times 10^{-18} \text{ Js/m}^3\). Frequency, \(\nu = 3 \times 10^{14}\) Hz. Speed of light, \(c = 3 \times 10^8\) m/s. Substitute these values into the conversion formula: \[ \rho_T(\lambda) = (8 \times 10^{-18} \text{ Js/m}^3) \times \frac{(3 \times 10^{14} \text{ s}^{-1})^2}{3 \times 10^8 \text{ m/s}} \] \[ \rho_T(\lambda) = (8 \times 10^{-18}) \times \frac{9 \times 10^{28}}{3 \times 10^8} \frac{\text{Js}}{\text{m}^3} \frac{\text{s}^{-2}}{\text{m/s}} \] \[ \rho_T(\lambda) = (8 \times 10^{-18}) \times (3 \times 10^{20}) \text{ J/m}^4 \] \[ \rho_T(\lambda) = 24 \times 10^2 \text{ J/m}^4 \] Step 4: Final Answer: 
The value of the energy density \(\rho_T(\lambda)\) is \(24 \times 10^2 \text{ J/m}^4\). The integer to be filled in is 24.