A hot body having the surface temperture `1127^(@)C` Determine the wavelength at wchich it radiates maximum energy . Given , Wien' s constant `=2.9xx10^(-3)` mK .

Options

A. `207.1Å`
B. `220.1Å`
C. `300.1Å`
D. `250.1Å`

Explanation

To determine the wavelength at which a hot body radiates maximum energy, we can use Wien's Displacement Law, which states:

\[ \lambda_{m} = \frac{b}{T} \]

where:
- \(\lambda_{m}\) is the wavelength at which maximum energy is radiated,
- \(b\) is Wien's constant (given as \(2.9 \times 10^{-3} \, \text{m K}\)),
- \(T\) is the absolute temperature in Kelvin.

### Step 1: Convert the temperature from Celsius to Kelvin
The surface temperature of the hot body is given as \(1127^{\circ}C\). To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273 \]

Substituting the given temperature:

\[ T = 1127 + 273 = 1400 \, K \]

### Step 2: Use Wien's Displacement Law to find the wavelength
Now that we have the temperature in Kelvin, we can substitute \(T\) and \(b\) into the formula for \(\lambda_{m}\):

\[ \lambda_{m} = \frac{b}{T} = \frac{2.9 \times 10^{-3} \, \text{m K}}{1400 \, K} \]

### Step 3: Calculate \(\lambda_{m}\)
Perform the calculation:

\[ \lambda_{m} = \frac{2.9 \times 10^{-3}}{1400} = 2.07142857 \times 10^{-6} \, m \]

To express this in Angstroms (1 Angstrom = \(10^{-10}\) m):

\[ \lambda_{m} = 2.07142857 \times 10^{-6} \, m \times \frac{10^{10} \, \text{Å}}{1 \, m} = 207.142857 \, \text{Å} \]

### Final Result
Thus, the wavelength at which the hot body radiates maximum energy is approximately:

\[ \lambda_{m} \approx 207.1 \, \text{Å} \]

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