Understanding Black Body Radiation and Wien's Displacement Law

A black body is an idealized object that absorbs all electromagnetic radiation incident on it and, when in thermal equilibrium, emits thermal radiation according to Planck's law. The distribution of energy emitted by a black body at a certain temperature follows a specific spectrum.

One important characteristic of this spectrum is that it has a peak at a particular wavelength. This peak wavelength shifts as the temperature of the black body changes. This relationship is described by Wien's Displacement Law.

Wien's Displacement Law

Wien's Displacement Law states that the wavelength at which the emission of a black body is maximum, denoted by \(\lambda_{max}\), is inversely proportional to its absolute temperature \(T\).

Mathematically, this can be written as:

\[ \lambda_{max} \cdot T = b \]

where \(b\) is Wien's displacement constant, approximately \(2.898 \times 10^{-3} \text{ m} \cdot \text{K}\).

This law implies that as the temperature of a black body increases, the peak wavelength shifts towards shorter wavelengths (towards the blue or ultraviolet end of the spectrum), and as the temperature decreases, the peak wavelength shifts towards longer wavelengths (towards the red or infrared end).

Applying Wien's Law to the Problem

We are given the initial temperature and the corresponding maximum emission wavelength for a black body, and we need to find the temperature at which it emits maximum energy at a different wavelength. Since the object is the same black body, the constant \(b\) in Wien's law remains constant.

Let \(T_1\) be the initial temperature and \(\lambda_{max,1}\) be the initial maximum emission wavelength. Let \(T_2\) be the final temperature and \(\lambda_{max,2}\) be the final maximum emission wavelength.

According to Wien's Displacement Law:

\[ \lambda_{max,1} \cdot T_1 = b \]

\[ \lambda_{max,2} \cdot T_2 = b \]

Since both expressions equal the same constant \(b\), we can equate them:

\[ \lambda_{max,1} \cdot T_1 = \lambda_{max,2} \cdot T_2 \]

Given Values

• Initial temperature, \(T_1 = 2000 \text{ K}\)
• Initial maximum wavelength, \(\lambda_{max,1} = 1.56 \text{ µm}\)
• Final maximum wavelength, \(\lambda_{max,2} = 1.8 \text{ µm}\)

Calculating the Final Temperature

We need to find \(T_2\). We can rearrange the equation:

\[ T_2 = \frac{\lambda_{max,1} \cdot T_1}{\lambda_{max,2}} \]

Now, substitute the given values into the equation:

\[ T_2 = \frac{(1.56 \text{ µm}) \cdot (2000 \text{ K})}{1.8 \text{ µm}} \]

Notice that the units of wavelength (\(\text{µm}\)) cancel out, leaving the unit of temperature (\(\text{K}\)), which is what we expect.

\[ T_2 = \frac{1.56 \times 2000}{1.8} \text{ K} \]

\[ T_2 = \frac{3120}{1.8} \text{ K} \]

Performing the division:

\[ T_2 \approx 1733.33 \text{ K} \]

Conclusion

The temperature at which the black body will emit maximum energy at a wavelength of 1.8 µm is approximately 1733 K.

Comparing this result with the given options:

• 1153 K
• 1353 K
• 1733 K
• 1533 K

The calculated value is closest to 1733 K.