Explanation Given data: Wavelength of the light is 550 nm. Lasting time of the flash is 100 ms. Power of the flash is 1.2 mW. Formula used: Write the expression for energy and wavelength. E photon = h c λ (1) Here, E photon is the energy of the photon, h is the plank’s constant with value 6.63 × 10 − 34 J ⋅ s , c is the speed of light with value 3.0 × 10 8 m s , and λ is the wavelength. Write the expression for power. P = E t (2) Here, P is the power, E is the energy, and t is the time. Explanation: Substitute 6.63 × 10 − 34 J ⋅ s for h , 3.0 × 10 8 m s for c and 550 × 10 − 9 m for λ in equation (1), E photon = ( 6.63 × 10 − 34 J ⋅ s ) ( 3.0 × 10 8 m s ) 550 × 10 − 9 m = ( 6.63 × 10 − 34 J ) ( 3.0 × 10 8 ) 550 × 10 − 9 Consider conversion factor for J to eV. 1 J = 1 1.6 × 10 − 19 eV Convert the energy value to eV by using conversion factor

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ISBN: 9780134568430

Author: Knight

Publisher: PEARSON

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Question

Chapter 28, Problem 22P

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The number of photons emitted in one flash.

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Chapter 28, Problem 21P

Chapter 28, Problem 23P

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The number of photons emitted in one flash.

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