Chapter 38: Q1Q (page 1180)

Photon A has twice the energy of photon B. (a) Is the momentum of A less than, equal to, or greater than that of B? (b) Is the wavelength of A less than, equal to, or greater than that of B?

Short Answer

Expert verified

a) The momentum of A is greater than the momentum of B.

b) The wavelength of A is less than the wavelength of B.

Step by step solution

Introduction

Monochromatic lights are single-wavelength light, where mono refers to single, and chroma means colour. Visible light of a narrow band of wavelengths is classified as monochromatic lights. It features a wavelength within a short wavelength range.

(a) Determine the greatest wavelength of light

Here, $E_{A} = 2 E_{B}$

But the momentum of the photon $p = \frac{E}{c}$

E is the Energy of the photon

c is the speed of the light (which is a constant).

As $E_{A} = 2 E_{B}$

$\begin{array}{l} \Rightarrow E_{A} > E_{a} \\ \Rightarrow \frac{E_{A}}{c} > \frac{E_{B}}{c} \\ \Rightarrow p_{A} > p_{B} \end{array}$

Therefore, the momentum of A is greater than the momentum of B.

(b) Determine region of the electromagnetic spectrum

Here, $E_{A} = 2 E_{B}$

localid="1663058432865" $\Rightarrow E_{A} > E_{B}$

But energy $E = h v$ , where h is Planck's constant

Frequency of light, $ν = \frac{h c}{λ}$

$∴ E = \frac{h c}{λ} E_{A} > E_{B} \Rightarrow \frac{h c}{λ_{A}} > \frac{h c}{λ_{B}} \Rightarrow λ_{A} < λ_{B}$

Therefore, wavelength of A is less than the wavelength of B.

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Photon A has twice the energy of photon B. (a) Is the momentum of A less than, equal to, or greater than that of B? (b) Is the wavelength of A less than, equal to, or greater than that of B?

Short Answer

Step by step solution

Introduction

(a) Determine the greatest wavelength of light

(b) Determine region of the electromagnetic spectrum

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