Explanation Given data : Consider that the wavelength of visible light is 500 nm. Explanation: Consider the expression for the energy of the photon. E photon = h f (1) Here, h is Planck's constant, which is generally known as 6.63 × 10 − 34 J ⋅ s . Write the expression for the frequency of the photon f . f = c λ (2) Here, λ is wavelength of photon, which is given as 500 nm, and c is the velocity of light in vacuum, which is generally 3 × 10 8 m s . Substitute equation (2) in (1). E photon = h ( c λ ) Substitute 6.63 × 10 − 34 J ⋅ s for h , 3 × 10 8 m s for c , and 500 nm for λ , E photon = ( 6.63 × 10 − 34 J ⋅ s ) ( 3 × 10 8 m s 500 nm ) = ( 6.63 × 10 − 34 J ⋅ s ) ( 3 × 10 8 m s 500 × 10 − 9 m ) = 3

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ISBN: 9780321989246

Author: Knight

Publisher: PEARSON CO

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Nature of light

Light is transverse and electromagnetic radiation that exhibits the properties of a wave as well as a particle. For this reason, light is said to have a “dual nature.”

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Chapter 25, Problem 40P

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To find: The energy of visible light’s photon.

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Chapter 25, Problem 39P

Chapter 25, Problem 41P

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Chapter 25 Solutions

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Chapter 25 Solutions