A black body at `1373 ^@C` emits maximum energy corresponding to a wavelength of 1.78 microns. The temperature of the moon for which `lamda_m=14` micron wood be

Options

A. `62.6 ^@C`
B. `-58.9^@C`
C. `-63.7 ^@C`
D. `64.2 ^@C`

Answer

To solve the problem, we will use Wien's Displacement Law, which states that the product of the maximum wavelength (\( \lambda_m \)) and the absolute temperature (T) of a black body is a constant. The formula is given by:

\[ \lambda_m \cdot T = b \]

where \( b \) is Wien's displacement constant, approximately equal to \( 2898 \, \mu m \cdot K \).

### Step 1: Convert the temperature from Celsius to Kelvin
The temperature of the black body is given as \( 1373 \, ^\circ C \). To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

Calculating:

\[ T_1 = 1373 + 273.15 = 1646.15 \, K \]

### Step 2: Use Wien's Law for the first black body
We know the maximum wavelength \( \lambda_{m1} = 1.78 \, \mu m \) and the corresponding temperature \( T_1 = 1646.15 \, K \). Using Wien's Law:

\[ \lambda_{m1} \cdot T_1 = b \]

Calculating \( b \):

\[ b = 1.78 \, \mu m \cdot 1646.15 \, K = 2929.88 \, \mu m \cdot K \]

### Step 3: Set up the equation for the moon's temperature
Now, we need to find the temperature \( T_2 \) for the moon where the maximum wavelength \( \lambda_{m2} = 14 \, \mu m \).

Using Wien's Law again:

\[ \lambda_{m2} \cdot T_2 = b \]

Substituting the known values:

\[ 14 \, \mu m \cdot T_2 = 2929.88 \, \mu m \cdot K \]

### Step 4: Solve for \( T_2 \)
Rearranging the equation to solve for \( T_2 \):

\[ T_2 = \frac{2929.88 \, \mu m \cdot K}{14 \, \mu m} \]

Calculating:

\[ T_2 = 209.27 \, K \]

### Step 5: Convert \( T_2 \) from Kelvin to Celsius
To convert from Kelvin back to Celsius:

\[ T(°C) = T(K) - 273.15 \]

Calculating:

\[ T_2(°C) = 209.27 - 273.15 = -63.88 \, ^\circ C \]

### Final Answer
The temperature of the moon corresponding to a maximum wavelength of \( 14 \, \mu m \) is approximately \( -63.88 \, ^\circ C \).

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