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Consider the Earth to be a blackbody with an average temperature of $15^\circ\text{C}$. Find out the wavelength ($\lambda_{max}$) at which maximum power is radiated :

This question was previously asked in
CUET PG 2026 Agri-Business Management Question Paper (25-Mar-2026) (Shift 2)
The correct answer is
$\sim 10.1 \mu\text{m}$

Calculating Earth's Maximum Radiation Wavelength

To find the wavelength ($\lambda_{max}$) at which maximum power is radiated by a blackbody, we use Wien's Displacement Law. The law states:

$ \lambda_{max} T = b $

Where:

  • $\lambda_{max}$ is the wavelength of maximum radiation.
  • $T$ is the absolute temperature in Kelvin (K).
  • $b$ is Wien's displacement constant, approximately $2.898 \times 10^{-3} \text{ m}\cdot\text{K}$.

Step 1: Convert Temperature to Kelvin

The given temperature is $15^\circ\text{C}$. We need to convert this to Kelvin:

$ T(\text{K}) = T(^\circ\text{C}) + 273.15 $ $ T = 15 + 273.15 = 288.15 \text{ K} $

Step 2: Apply Wien's Displacement Law

Now, substitute the values into Wien's formula to find $\lambda_{max}$:

$ \lambda_{max} = \frac{b}{T} $ $ \lambda_{max} = \frac{2.898 \times 10^{-3} \text{ m}\cdot\text{K}}{288.15 \text{ K}} $ $ \lambda_{max} \approx 1.0057 \times 10^{-5} \text{ m} $

Step 3: Convert Wavelength to Micrometers

The options are given in micrometers ($\mu\text{m}$). Convert the calculated wavelength:

Since $1 \text{ m} = 10^6 \mu\text{m}$:

$ \lambda_{max} \approx (1.0057 \times 10^{-5} \text{ m}) \times (10^6 \mu\text{m}/\text{m}) $ $ \lambda_{max} \approx 10.057 \mu\text{m} $

Conclusion

The calculated wavelength is approximately $10.1 \mu\text{m}$. This corresponds to the wavelength at which the Earth, considered as a blackbody at $15^\circ\text{C}$, radiates maximum power.

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