Calculate the value of `lamda_("max")` for radiation from a body having surface temperature 3000 K.

Options

A. 9935 Å
B. 9656 Å
C. 9421 Å
D. 9178 Å

Answer

To calculate the value of `λ_max` (the maximum wavelength) for radiation from a body with a surface temperature of 3000 K, we can use Wien's Displacement Law. Here’s the step-by-step solution:

### Step 1: Understand Wien's Displacement Law
Wien's Displacement Law states that the wavelength at which the emission of a black body spectrum is maximized (λ_max) is inversely proportional to the absolute temperature (T) of the body. The law can be expressed mathematically as:
\[ \lambda_{\text{max}} \cdot T = b \]
where \( b \) is Wien's displacement constant, approximately equal to \( 2.9 \times 10^{-3} \, \text{m} \cdot \text{K} \).

### Step 2: Rearrange the formula to find λ_max
From the equation above, we can rearrange it to solve for \( \lambda_{\text{max}} \):
\[ \lambda_{\text{max}} = \frac{b}{T} \]

### Step 3: Substitute the values
Now, substitute the value of \( b \) and the given temperature \( T = 3000 \, \text{K} \):
\[ \lambda_{\text{max}} = \frac{2.9 \times 10^{-3} \, \text{m} \cdot \text{K}}{3000 \, \text{K}} \]

### Step 4: Perform the calculation
Calculating the above expression:
\[ \lambda_{\text{max}} = \frac{2.9 \times 10^{-3}}{3000} \]
\[ \lambda_{\text{max}} = 9.66667 \times 10^{-7} \, \text{m} \]

### Step 5: Convert to Angstroms
Since 1 meter = \( 10^{10} \) Angstroms, we can convert \( \lambda_{\text{max}} \) to Angstroms:
\[ \lambda_{\text{max}} = 9.66667 \times 10^{-7} \, \text{m} \times 10^{10} \, \text{Å/m} \]
\[ \lambda_{\text{max}} = 9666.67 \, \text{Å} \]

### Final Answer
Thus, the value of \( \lambda_{\text{max}} \) for radiation from a body having a surface temperature of 3000 K is approximately:
\[ \lambda_{\text{max}} \approx 9666.67 \, \text{Å} \]