Question

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

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, $E_A=2E_B$

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 $E_A=2E_B$

$\begin{array}{l}\Rightarrow E_A>E_a\\ \Rightarrow \frac{E_A}{c}>\frac{E_B}{c}\\ \Rightarrow p_A>p_B\end{array}$

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

(b) Determine region of the electromagnetic spectrum

Here, $E_A=2E_B$

localid="1663058432865" $\Rightarrow E_A>E_B$

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

Frequency of light, $\nu =\frac{hc}{\lambda }$

$$\begin{gathered} \therefore E=\frac{hc}{\lambda } \\ {E}_A>E_B\Rightarrow \frac{hc}{{\lambda }_A}>\frac{hc}{{\lambda }_B} \\ \Rightarrow {\lambda }_A<{\lambda }_B \end{gathered}$$

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