Consider a parallel beam of light of wavelength 600 nm and intensity 100 W/m2 .

(a) find the energy and linear momentum of each photon.

(b) How many photons cross 1 cm2 area perpendicular to the beam in one second?

To tackle your question about the properties of a parallel beam of light, we’ll break it down into two parts: first, calculating the energy and linear momentum of each photon, and then determining how many photons cross a specific area in a given time. Let’s dive in!

Calculating Energy and Linear Momentum of Each Photon

The energy of a photon can be calculated using the formula:

E = h * f

Where:

• E = energy of the photon (in joules)
• h = Planck's constant, approximately 6.626 x 10^-34 J·s
• f = frequency of the light (in hertz)

To find the frequency (f), we can use the relationship between the speed of light (c), wavelength (λ), and frequency:

f = c / λ

Here, the speed of light (c) is about 3 x 10^8 m/s, and the wavelength (λ) is given as 600 nm, which we need to convert to meters:

600 nm = 600 x 10^-9 m

Now, substituting the values into the frequency formula:

f = (3 x 10^8 m/s) / (600 x 10^-9 m) = 5 x 10^14 Hz

Now that we have the frequency, we can calculate the energy:

E = (6.626 x 10^-34 J·s) * (5 x 10^14 Hz) = 3.313 x 10^-19 J

Finding Linear Momentum of Each Photon

The linear momentum (p) of a photon is given by the formula:

p = E / c

Substituting the energy we just calculated:

p = (3.313 x 10^-19 J) / (3 x 10^8 m/s) = 1.104 x 10^-27 kg·m/s

Determining the Number of Photons Crossing an Area

Next, let’s find out how many photons cross a 1 cm² area perpendicular to the beam in one second. First, we need to calculate the total power (P) of the beam:

P = intensity * area

Given that the intensity is 100 W/m² and the area is 1 cm² (which is 0.0001 m²), we can calculate:

P = 100 W/m² * 0.0001 m² = 0.01 W

Now, we know that power is also the energy per unit time:

P = E_photon * number_of_photons / time

Rearranging this gives us:

number_of_photons = P * time / E_photon

Assuming time is 1 second, we can substitute the values:

number_of_photons = (0.01 W * 1 s) / (3.313 x 10^-19 J) ≈ 3.01 x 10^16 photons

Summary of Results

To summarize:

• The energy of each photon is approximately 3.313 x 10^-19 J.
• The linear momentum of each photon is about 1.104 x 10^-27 kg·m/s.
• Approximately 3.01 x 10^16 photons cross a 1 cm² area perpendicular to the beam in one second.

This analysis shows how light behaves at the quantum level, illustrating the relationship between energy, momentum, and the number of photons in a beam. If you have any further questions or need clarification on any of these points, feel free to ask!