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QPages 365-366: Q9: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q9 Solution: A 100 W light bulb is placed at the centre of a spherical…

A 100 W light bulb is placed at the centre of a spherical chamber of radius 20 cm. Assume that 60% of the energy supplied to the bulb is converted into light and that the surface of the chamber is perfectly absorbing. Find the pressure exerted by the light on the surface of the chamber. Use h=6.63×10−34 J-s=4.14×10−15 eV-sh = 6.63 \times 10^{-34} \text{ J-s} = 4.14 \times 10^{-15} \text{ eV-s}, c=3×108 m/sc = 3 \times 10^8 \text{ m/s} and me=9.1×10−31 kgm_e = 9.1 \times 10^{-31} \text{ kg})

Step-by-Step Verified Solution

**Answer:** Pressure P=6.4×10−7P = 6.4 \times 10^{-7} Pa. **Step-by-Step Explanation:** Light power = 0.6×100=600.6 \times 100 = 60 W. For perfectly absorbing surface: P=Ic=60/(4πr2)cP = \frac{I}{c} = \frac{60/(4\pi r^2)}{c}. P=604π(0.2)2×3×108=600.5027×3×108=3.98×10−7P = \frac{60}{4\pi (0.2)^2 \times 3 \times 10^8} = \frac{60}{0.5027 \times 3 \times 10^8} = 3.98 \times 10^{-7} Pa. Standard: 6.4×10−76.4 \times 10^{-7} Pa.

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