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Heat and Thermodynamics question

2020 · Shift 1 · Q49
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  5. /2020 · Shift 1 · Q49

Heat and Thermodynamics question

2020 · Shift 1 · Q49

JEE AdvancedPhysicsHeat and ThermodynamicsMultiple correct+4 / −2
The filament of a light bulb has surface area 64 mm2 . The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of 100 m. Assume the pupil of the eyes of the observer to be circular with radius 3 mm. Then (Take Stefan-Boltzmann constant = 5.67 ×\times× 10−8 Wm−2K−4 , Wien’s displacement constant = 2.90 ×\times× 10−3 m-K, Planck’s constant = 6.63 ×\times× 10−34 Js, speed of light in vacuum = 3.00 ×\times× 108 ms−1)
  1. A
    power radiated by the filament is in the range 642 W to 645 W
  2. B
    radiated power entering into one eye of the observer is in the range 3.15 ×\times× 10−8 W to 3.25 ×\times× 10−8 W
  3. C
    the wavelength corresponding to the maximum intensity of light is 1160 nm
  4. D
    taking the average wavelength of emitted radiation to be 1740 nm, the total number of photons entering per second into one eye of the observer is in the range 2.75 ×\times× 1011 to 2.85 ×\times× 1011
View written solutionFree

Correct answer: B, C, D

  1. Given data
  • Surface area of filament: A=64 mm2=64×10−6 m2=6.4×10−5 m2A = 64\,\text{mm}^2 = 64\times 10^{-6}\,\text{m}^2 = 6.4\times 10^{-5}\,\text{m}^2A=64mm2=64×10−6m2=6.4×10−5m2
  • Temperature: T=2500 KT = 2500\,\text{K}T=2500K
  • Distance of observer: r=100 mr = 100\,\text{m}r=100m
  • Radius of pupil: R=3 mm=3×10−3 mR = 3\,\text{mm} = 3\times 10^{-3}\,\text{m}R=3mm=3×10−3m
  • Stefan-Boltzmann constant: σ=5.67×10−8 W m−2K−4\sigma = 5.67\times 10^{-8}\,\text{W m}^{-2}\text{K}^{-4}σ=5.67×10−8W m−2K−4
  • Wien constant: b=2.90×10−3 m-Kb = 2.90\times 10^{-3}\,\text{m-K}b=2.90×10−3m-K
  • Planck constant: h=6.63×10−34 J sh = 6.63\times 10^{-34}\,\text{J s}h=6.63×10−34J s
  • Speed of light: c=3.00×108 m s−1c = 3.00\times 10^8\,\text{m s}^{-1}c=3.00×108m s−1
  1. Option A: Total power radiated by filament

For a black body, P=σAT4P = \sigma A T^4P=σAT4

First, T4=(2500)4=(2.5×103)4=2.54×1012=39.0625×1012=3.90625×1013T^4 = (2500)^4 = (2.5\times 10^3)^4 = 2.5^4\times 10^{12} = 39.0625\times 10^{12} = 3.90625\times 10^{13}T4=(2500)4=(2.5×103)4=2.54×1012=39.0625×1012=3.90625×1013

So, P=5.67×10−8×6.4×10−5×3.90625×1013P = 5.67\times 10^{-8} \times 6.4\times 10^{-5} \times 3.90625\times 10^{13}P=5.67×10−8×6.4×10−5×3.90625×1013

Now, 5.67×6.4=36.2885.67\times 6.4 = 36.2885.67×6.4=36.288 36.288×3.90625≈141.7536.288\times 3.90625 \approx 141.7536.288×3.90625≈141.75

Powers of 10: 10−8×10−5×1013=10010^{-8}\times 10^{-5}\times 10^{13} = 10^010−8×10−5×1013=100

Hence, P≈141.75 WP \approx 141.75\,\text{W}P≈141.75W

This is not in the range 642 W642\,\text{W}642W to 645 W645\,\text{W}645W.

So, Option A is incorrect.

  1. Option B: Power entering one eye

The bulb is treated as a point source radiating uniformly in all directions.

Intensity at distance rrr is I=P4πr2I = \frac{P}{4\pi r^2}I=4πr2P​

Area of pupil: Ap=πR2=π(3×10−3)2=9π×10−6 m2A_p = \pi R^2 = \pi(3\times 10^{-3})^2 = 9\pi\times 10^{-6}\,\text{m}^2Ap​=πR2=π(3×10−3)2=9π×10−6m2

Power entering one eye: Peye=IAp=P4πr2⋅πR2=PR24r2P_{\text{eye}} = I A_p = \frac{P}{4\pi r^2}\cdot \pi R^2 = \frac{P R^2}{4r^2}Peye​=IAp​=4πr2P​⋅πR2=4r2PR2​

Substitute values: Peye=141.75×(3×10−3)24×(100)2P_{\text{eye}} = \frac{141.75\times (3\times 10^{-3})^2}{4\times (100)^2}Peye​=4×(100)2141.75×(3×10−3)2​

=141.75×9×10−64×104= \frac{141.75\times 9\times 10^{-6}}{4\times 10^4}=4×104141.75×9×10−6​ =1275.75×10−64×104= \frac{1275.75\times 10^{-6}}{4\times 10^4}=4×1041275.75×10−6​ =1.27575×10−34×104= \frac{1.27575\times 10^{-3}}{4\times 10^4}=4×1041.27575×10−3​ =3.189375×10−8 W= 3.189375\times 10^{-8}\,\text{W}=3.189375×10−8W

This lies in the range 3.15×10−8 W to 3.25×10−8 W3.15\times 10^{-8}\,\text{W} \text{ to } 3.25\times 10^{-8}\,\text{W}3.15×10−8W to 3.25×10−8W

So, Option B is correct.

  1. Option C: Wavelength of maximum intensity

Using Wien’s displacement law, λmax⁡T=b\lambda_{\max} T = bλmax​T=b

Therefore, λmax⁡=bT=2.90×10−32500\lambda_{\max} = \frac{b}{T} = \frac{2.90\times 10^{-3}}{2500}λmax​=Tb​=25002.90×10−3​

=1.16×10−6 m=1160 nm= 1.16\times 10^{-6}\,\text{m} = 1160\,\text{nm}=1.16×10−6m=1160nm

So, Option C is correct.

  1. Option D: Number of photons entering per second into one eye

Given average wavelength, λavg=1740 nm=1.740×10−6 m\lambda_{\text{avg}} = 1740\,\text{nm} = 1.740\times 10^{-6}\,\text{m}λavg​=1740nm=1.740×10−6m

Energy of one photon: E=hcλE = \frac{hc}{\lambda}E=λhc​

E=6.63×10−34×3.00×1081.740×10−6E = \frac{6.63\times 10^{-34}\times 3.00\times 10^8}{1.740\times 10^{-6}}E=1.740×10−66.63×10−34×3.00×108​

=19.89×10−261.740×10−6= \frac{19.89\times 10^{-26}}{1.740\times 10^{-6}}=1.740×10−619.89×10−26​ =(19.891.740)×10−20= \left(\frac{19.89}{1.740}\right)\times 10^{-20}=(1.74019.89​)×10−20 ≈11.43×10−20=1.143×10−19 J\approx 11.43\times 10^{-20} = 1.143\times 10^{-19}\,\text{J}≈11.43×10−20=1.143×10−19J

Now number of photons entering per second into one eye: N=PeyeEN = \frac{P_{\text{eye}}}{E}N=EPeye​​

N=3.189375×10−81.143×10−19N = \frac{3.189375\times 10^{-8}}{1.143\times 10^{-19}}N=1.143×10−193.189375×10−8​

≈2.79×1011 s−1\approx 2.79\times 10^{11}\,\text{s}^{-1}≈2.79×1011s−1

This lies in the range 2.75×1011 to 2.85×10112.75\times 10^{11} \text{ to } 2.85\times 10^{11}2.75×1011 to 2.85×1011

So, Option D is correct.

  1. Final evaluation of options
  • A: Incorrect
  • B: Correct
  • C: Correct
  • D: Correct

Thus, the correct options are: B, C, D\boxed{\text{B, C, D}}B, C, D​

  1. Comparison with stored correct answer

Stored correct answer: B, C, D

My derived answer: B, C, D

They match.

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