Problem 78 - Master Resource Book in JEE Main Physics

Question

Three discs \(A, B\) and \(C\) having radii \(2 \mathrm{~m}, 4 \mathrm{~m}\) and \(6 \mathrm{~m}\) respectively are coated with carbon black on their other surfaces. The wavelengths corresponding to maximum intensity are \(300 \mathrm{~nm}, 400 \mathrm{~nm}\) and \(500 \mathrm{~nm}\) respectively. The power radiated by them are \(Q_{a}, Q_{b}\) and \(Q_{c}\) respectively (a) \(Q_{a}\) is maximum (b) \(Q_{b}\) is maximum (c) \(Q_{c}\) is maximum (d) \(Q_{a}=Q_{b}=Q_{c}\)

Step-by-Step Solution

Answer

\(Q_c\) is maximum.

Step 1: Identify the given quantities

We have three discs with different radii: \( r_A = 2\, \text{m} \), \( r_B = 4\, \text{m} \), and \( r_C = 6\, \text{m} \). The wavelengths of maximum intensity are \( \lambda_A = 300\, \text{nm} \), \( \lambda_B = 400\, \text{nm} \), and \( \lambda_C = 500\, \text{nm} \). We need to find which disc radiates maximum power: \( Q_a, Q_b \text{ or } Q_c \).

Step 2: Apply Wien's Displacement Law

Wien’s displacement law states that the wavelength of maximum emission \( \lambda_m \) for a blackbody is inversely proportional to its temperature \( T \), given as \( \lambda_m T = b \) where \( b \approx 2.898 \times 10^{-3} \text{ m} \cdot \text{K} \). This implies that the temperature \( T_A = \frac{b}{\lambda_A} \), \( T_B = \frac{b}{\lambda_B} \), and \( T_C = \frac{b}{\lambda_C} \).

Step 3: Calculate Temperatures

Using Wien's law, we find:\[ T_A = \frac{2.898 \times 10^{-3}}{300 \times 10^{-9}} = \frac{2.898}{0.3} \times 10^4 = 9660\, \text{K} \]\[ T_B = \frac{2.898 \times 10^{-3}}{400 \times 10^{-9}} = \frac{2.898}{0.4} \times 10^4 = 7245\, \text{K} \]\[ T_C = \frac{2.898 \times 10^{-3}}{500 \times 10^{-9}} = \frac{2.898}{0.5} \times 10^4 = 5796\, \text{K} \]

Step 4: Use Stefan-Boltzmann Law to Determine Power Radiated

The power radiated by a black body is given by the Stefan-Boltzmann law:\[ Q = \sigma A T^4 \]where \( \sigma \) is the Stefan-Boltzmann constant, and \( A \) is the surface area. For discs,\( A_A = \pi r_A^2 \), \( A_B = \pi r_B^2 \), and \( A_C = \pi r_C^2 \).

Step 5: Calculate Areas of Discs

Calculate the areas:\[ A_A = \pi (2)^2 = 4\pi \]\[ A_B = \pi (4)^2 = 16\pi \]\[ A_C = \pi (6)^2 = 36\pi \]

Step 6: Compute Radiated Powers

Now, compute the power.\[ Q_A = \sigma 4\pi (9660)^4 \]\[ Q_B = \sigma 16\pi (7245)^4 \]\[ Q_C = \sigma 36\pi (5796)^4 \]We see that although \( T_A \) is highest, \( Q_C \) will be maximum due to the combination of larger area and significant temperature contribution.

Key Concepts

Wien's Displacement Law

Wien's Displacement Law is a fundamental principle in understanding blackbody radiation. It reveals that the wavelength at which a blackbody emits maximum radiation, denoted as \( \lambda_m \), is inversely related to its temperature \( T \). This is mathematically expressed as: \[ \lambda_m T = b \] where \( b \approx 2.898 \times 10^{-3} \, \text{m} \cdot \text{K} \). This relationship implies that higher temperature bodies emit radiation at shorter wavelengths.

• As an object's temperature increases, the peak of its emitted radiation shifts to shorter wavelengths.
• This law allows us to determine the temperature of an object by observing the peak wavelength of its emitted radiation.

For example, in our exercise, the temperatures of discs \(A\), \(B\), and \(C\) were calculated using their respective peak wavelengths: 300 nm, 400 nm, and 500 nm. By plugging these values into the Wien's Law formula, we determined that Disc \(A\) had the highest temperature, followed by \(B\), and finally \(C\). This concept helps us connect visible observations to thermal properties.

Blackbody Radiation

Blackbody radiation is the theoretical emission of electromagnetic radiation from an idealized object called a blackbody. A blackbody is an ideal emitter and absorber of energy.

• All bodies emit radiation, but a blackbody is a perfect model that absorbs all incident radiation without reflecting any.
• The amount and type of radiation emitted depends on the body's temperature.

In reality, no perfect blackbodies exist, but many objects approximate the behavior. The Stefan-Boltzmann Law helps in quantifying this radiation by stating that the power radiated per unit area of a blackbody \( Q \) is directly proportional to the fourth power of its temperature \( T \). This is represented as: \[ Q = \sigma A T^4 \] where \( \sigma \) is the Stefan-Boltzmann constant, and \( A \) is the surface area of the emitting body. In the exercise, discs were considered as blackbodies with their radiating power determined by combining their surface areas and temperatures calculated from Wien's Law.

Surface Area Calculation

Surface area is essential to determine how much energy is radiated by a blackbody. The surface area affects the total power radiated by a body when combined with its temperature. For circular discs, the area \( A \) is given by the formula: \[ A = \pi r^2 \]

• A larger surface area means more space over which radiation can be emitted.
• In scenarios involving blackbody radiation, this becomes a crucial factor in determining total emitted energy.

In the exercise, the areas of the discs \(A\), \(B\), and \(C\) were calculated using their radii: 2 m, 4 m, and 6 m, respectively. Thus, the surface areas were found to be \(4\pi\), \(16\pi\), and \(36\pi\). Despite a lower temperature, disc \(C\)'s larger area resulted in it radiating more energy than the other two. Thus, both the temperature and the area combine to dictate the total energy emitted by each disc.