Explanation Given data: Wavelength of the radiation is 680 nm. Formula used: Write the expression for energy of the photon. E photon = h f (1) Here, E photon is the energy of the photon, h is the plank’s constant with value as 6.63 × 10 − 34 J ⋅ s , and f is the frequency. Write the expression for frequency. f = c λ (2) Here, c is the speed of light with value 3 × 10 8 m s , and λ is the wavelength. Write the expression for power. P = Δ E Δ t (3) Here, P is the power dissipated, Δ E is the energy, and Δ t is the time period. Write the expression for number of photons. N = E E photon (4) Here, N is the number of photons and E photon is the energy of each photon. Write the expression for intensity and area. I = P A (5) Here, I is the intensity, P is the power, and A is ht area. Write the expression for area of the circle, A = π r 2 (6) Here, A is the area of the circle. r is the radius of the circle. Explanation: Combining equation (1) and (2), E photon = h c λ (7) Combining equation (5), (3) and (4), I = N E photon A Δ t Modifying the equation as follows. I = ( N Δ t ) × ( E photon A ) = ( N Δ t ) × ( h c λ A ) { ∵ E photon = h c λ } I = ( N Δ t ) × ( h c A λ ) For number of photons per second ( N Δ t ) , I = ( N Δ t ) × ( h c A λ ) ( N Δ t ) = I A λ h c (8) Area of the pupil is calculated as follows

College Physics: A Strategic Approach (4th Edition)

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ISBN: 9780134609034

Author: Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field

Publisher: PEARSON

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Chapter 25, Problem 49P

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Chapter 25 Solutions

College Physics: A Strategic Approach (4th Edition)

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Chapter 25 Solutions