JEE Challenger
More from Thermal Properties of Matter
PhyHardAdv 2020 • Paper 1

Blackbody Radiation and Observer Eye Photon Collection

A light bulb's filament features a surface area of 64 mm264 \text{ mm}^2. Treating the filament as a black body operating at a temperature of 2500 K2500 \text{ K}, it emits radiation as a point source when viewed from far away. An observer located at a distance of 100 m100 \text{ m} observes the bulb at night. Assume the observer's eye pupil is circular with a radius of 3 mm3 \text{ mm}.

Take:

  • Stefan-Boltzmann constant =5.67×10−8 W m−2 K−4= 5.67 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}
  • Wien's displacement constant =2.90×10−3 m K= 2.90 \times 10^{-3} \text{ m K}
  • Planck's constant =6.63×10−34 J s= 6.63 \times 10^{-34} \text{ J s}
  • Speed of light in vacuum =3.00×108 m s−1= 3.00 \times 10^8 \text{ m s}^{-1}

Select the correct statement(s):

Options

A

The power radiated by the filament is in the range 642 W642 \text{ W} to 645 W645 \text{ W}

B

The radiated power entering into one eye of the observer is in the range 3.15×10−8 W3.15 \times 10^{-8} \text{ W} to 3.25×10−8 W3.25 \times 10^{-8} \text{ W}

Correct
C

The wavelength corresponding to the peak intensity of light is 1160 nm1160 \text{ nm}

Correct
D

Taking the average wavelength of emitted radiation to be 1740 nm1740 \text{ nm}, the total number of photons entering per second into one eye of the observer lies in the range 2.75×10112.75 \times 10^{11} to 2.85×10112.85 \times 10^{11}

Correct

Topics & Concepts

Thermal Properties of MatterHeat Transfer

Step-by-Step Solution

To determine the correct statements, we analyze each option step-by-step using the principles of thermal radiation and quantum optics.

1. Calculation of Total Power Radiated by the Filament (Option A)

According to Stefan-Boltzmann Law, the total power PP radiated by a black body of surface area AA at absolute temperature TT is given by: P=σAT4P = \sigma A T^4

Given data:

  • σ=5.67×10−8 W m−2 K−4\sigma = 5.67 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}
  • A=64 mm2=64×10−6 m2A = 64 \text{ mm}^2 = 64 \times 10^{-6} \text{ m}^2
  • T=2500 KT = 2500 \text{ K}

Calculating T4T^4: T4=(2500)4=(2.5×103)4=39.0625×1012 K4T^4 = (2500)^4 = (2.5 \times 10^3)^4 = 39.0625 \times 10^{12} \text{ K}^4

Substituting these values into the expression for PP: P=(5.67×10−8)×(64×10−6)×(39.0625×1012)P = (5.67 \times 10^{-8}) \times (64 \times 10^{-6}) \times (39.0625 \times 10^{12}) P=5.67×64×39.0625×10−2 WP = 5.67 \times 64 \times 39.0625 \times 10^{-2} \text{ W}

Note that 64×39.0625=250064 \times 39.0625 = 2500, so: P=5.67×2500×10−2=5.67×25=141.75 WP = 5.67 \times 2500 \times 10^{-2} = 5.67 \times 25 = 141.75 \text{ W}

Since 141.75 W141.75 \text{ W} does not lie in the range 642 W642 \text{ W} to 645 W645 \text{ W}, Option A is incorrect.

2. Power Entering One Eye of the Observer (Option B)

The filament emits radiation isotropically as a point source. The power entering the eye pupil located at distance dd is: Peye=P×Apupil4πd2P_{\text{eye}} = P \times \frac{A_{\text{pupil}}}{4\pi d^2}

Given data:

  • d=100 md = 100 \text{ m}
  • r=3 mm=3×10−3 mr = 3 \text{ mm} = 3 \times 10^{-3} \text{ m}
  • Area of the pupil, Apupil=πr2=π(3×10−3)2=9π×10−6 m2A_{\text{pupil}} = \pi r^2 = \pi (3 \times 10^{-3})^2 = 9\pi \times 10^{-6} \text{ m}^2
  • Area of the sphere of radius dd, Atotal=4πd2=4π(100)2=4π×104 m2A_{\text{total}} = 4\pi d^2 = 4\pi (100)^2 = 4\pi \times 10^4 \text{ m}^2

Substituting the values: Peye=141.75×9π×10−64π×104=141.75×94×10−10P_{\text{eye}} = 141.75 \times \frac{9\pi \times 10^{-6}}{4\pi \times 10^4} = 141.75 \times \frac{9}{4} \times 10^{-10} Peye=141.75×2.25×10−10=318.9375×10−10 W=3.189375×10−8 WP_{\text{eye}} = 141.75 \times 2.25 \times 10^{-10} = 318.9375 \times 10^{-10} \text{ W} = 3.189375 \times 10^{-8} \text{ W}

Since 3.189375×10−8 W3.189375 \times 10^{-8} \text{ W} falls in the range 3.15×10−8 W3.15 \times 10^{-8} \text{ W} to 3.25×10−8 W3.25 \times 10^{-8} \text{ W}, Option B is correct.

3. Wavelength Corresponding to Peak Intensity (Option C)

By Wien's Displacement Law: λmT=b\lambda_m T = b

Where:

  • b=2.90×10−3 m Kb = 2.90 \times 10^{-3} \text{ m K}
  • T=2500 KT = 2500 \text{ K}

λm=bT=2.90×10−32500=1.16×10−6 m=1160 nm\lambda_m = \frac{b}{T} = \frac{2.90 \times 10^{-3}}{2500} = 1.16 \times 10^{-6} \text{ m} = 1160 \text{ nm}

Thus, Option C is correct.

4. Total Number of Photons Entering One Eye Per Second (Option D)

The energy of a single photon of average wavelength λ=1740 nm=1.74×10−6 m\lambda = 1740 \text{ nm} = 1.74 \times 10^{-6} \text{ m} is: E=hcλE = \frac{hc}{\lambda}

Given:

  • h=6.63×10−34 J sh = 6.63 \times 10^{-34} \text{ J s}
  • c=3.00×108 m s−1c = 3.00 \times 10^8 \text{ m s}^{-1}

E=6.63×10−34×3.00×1081.74×10−6=19.89×10−261.74×10−6=11.431×10−20 JE = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{1.74 \times 10^{-6}} = \frac{19.89 \times 10^{-26}}{1.74 \times 10^{-6}} = 11.431 \times 10^{-20} \text{ J}

The number of photons entering the eye per second, NN, is given by: N=PeyeE=3.189375×10−8 W11.431×10−20 J≈2.7899×1011 s−1N = \frac{P_{\text{eye}}}{E} = \frac{3.189375 \times 10^{-8} \text{ W}}{11.431 \times 10^{-20} \text{ J}} \approx 2.7899 \times 10^{11} \text{ s}^{-1}

Since 2.7899×10112.7899 \times 10^{11} lies in the range 2.75×10112.75 \times 10^{11} to 2.85×10112.85 \times 10^{11}, Option D is correct.

Conclusion

The correct options are B, C, and D.

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