A 100 W light bulb is positioned at the centre of a spherical chamber having a radius of 20 cm . Assuming that 60\% of the energy supplied to the bulb is converted into light and the chamber's surface is perfectly

Correct answer: A. 4.0 × 10^ -7 Pa Power of the light emitted, P = 60\% of 100 W = 60 W Radius of the spherical chamber, r = 20 cm = 0.2 m Intensity of light at the surface of the chamber is given by: I = (P)/(4 π r^2) For a perfectly absorbing surface, the radiation pressure p is: p = (I)/(c) = (P)/(4 π r^2 c) Substituting the given values: p = (60)/(4 π (0.2)^2 × 3 × 10^8) p = (60)/(4 π × 0.04 × 3 × 10^8) p = (60)/(0.48 π × 10^8) p = (125)/(π) × 10^ -8 Pa p 39.8 × 10^ -8 Pa = 3.98 × 10^ -7 Pa 4.0 × 10^ -7 Pa

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Photoelectric Effect and Wave–Particle Duality

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Photoelectric Effect and Wave–Particle Duality

A 100 W light bulb is positioned at the centre of a spherical...

100 \text{ W}

light bulb is positioned at the centre of a spherical chamber having a radius of

20 \text{ cm}

. Assuming that

60\%

of the energy supplied to the bulb is converted into light and the chamber's surface is perfectly absorbing, determine the pressure exerted by the light on the surface of the chamber.

Options

A $4.0 \times 10^{-7} \text{ Pa}$
B $8.0 \times 10^{-7} \text{ Pa}$
C $1.6 \times 10^{-6} \text{ Pa}$
D $6.6 \times 10^{-7} \text{ Pa}$

Correct answer

A. $4.0 \times 10^{-7} \text{ Pa}$

Step-by-step solution

Power of the light emitted,

P = 60\% \text{ of } 100 \text{ W} = 60 \text{ W}

Radius of the spherical chamber,

r = 20 \text{ cm} = 0.2 \text{ m}

Intensity of light at the surface of the chamber is given by:

I = \dfrac{P}{4 \pi r^2}

For a perfectly absorbing surface, the radiation pressure

p

is:

p = \dfrac{I}{c} = \dfrac{P}{4 \pi r^2 c}

Substituting the given values:

p = \dfrac{60}{4 \pi (0.2)^2 \times 3 \times 10^8}

p = \dfrac{60}{4 \pi \times 0.04 \times 3 \times 10^8}

p = \dfrac{60}{0.48 \pi \times 10^8}

p = \dfrac{125}{\pi} \times 10^{-8} \text{ Pa}

p \approx 39.8 \times 10^{-8} \text{ Pa} = 3.98 \times 10^{-7} \text{ Pa} \approx 4.0 \times 10^{-7} \text{ Pa}

Practice Photoelectric Effect and Wave–Particle Duality on Quantrex Academy →

More from Photoelectric Effect and Wave–Particle Duality