At noon on a clear day, sunlight reaches the earth's surface at Madison, Wisconsin, with an average power of approximately 4.00 kJ·s–1·m–2. If the sunlight consists of photons with an average wavelength of 510.0 nm, how many photons strike a 6.60 cm2 area per second?
DrBob your explanations are incredibly unclear.
To find the number of photons striking a given area per second, we need to calculate the photon flux.
1. Start by converting the power of sunlight to energy per second per square meter. Given that the sunlight power is 4.00 kJ·s–1·m–2, convert it to joules per second per square meter:
4.00 kJ·s–1·m–2 = 4.00 × 10^3 J·s–1·m–2.
2. Next, calculate the energy of an individual photon using the equation E = hc/λ, where E is the energy of a photon, h is Planck's constant (6.626 × 10^–34 J·s), c is the speed of light (3.00 × 10^8 m/s), and λ is the average wavelength of the photon. Plugging in the values gives:
E = (6.626 × 10^–34 J·s)(3.00 × 10^8 m/s) / (510.0 × 10^–9 m) = 3.88 × 10^–19 J.
3. Now, divide the power of sunlight per square meter by the energy of an individual photon to find the number of photons per second per square meter:
(4.00 × 10^3 J·s–1·m–2) / (3.88 × 10^–19 J) = 1.03 × 10^22 photons·s–1·m–2.
4. Finally, multiply the photon flux by the given area (6.60 cm^2) to calculate the number of photons striking the area per second:
(1.03 × 10^22 photons·s–1·m–2) × (6.60 × 10^–4 m^2) = 6.80 × 10^17 photons·s–1.
Therefore, approximately 6.80 × 10^17 photons strike a 6.60 cm^2 area per second.
To find the number of photons that strike a 6.60 cm² area per second, we need to consider the intensity of sunlight and the energy carried by individual photons.
First, let's convert the area in square meters:
1 cm² = (1/100)² m² = 0.0001 m²
So, the area is 6.60 cm² * 0.0001 m²/cm² = 0.00066 m².
Next, let's convert the average power from kilojoules per second per square meter (kJ·s⁻¹·m⁻²) to joules per second per square meter (J·s⁻¹·m⁻²):
1 kJ = 1000 J
Therefore, the average power is 4.00 kJ·s⁻¹·m⁻² * 1000 J/kJ = 4000 J·s⁻¹·m⁻².
Now, we need to calculate the energy carried by individual photons using Planck's equation:
E = hc/λ
Where:
- E is the energy of a photon
- h is Planck's constant (6.626 x 10⁻³⁴ J·s)
- c is the speed of light (approx. 3.00 x 10⁸ m/s)
- λ is the average wavelength of the sunlight (510.0 nm)
Converting the wavelength to meters:
510.0 nm = 510.0 x 10⁻⁹ m
Now, calculating the energy carried by each photon:
E = (6.626 x 10⁻³⁴ J·s * 3.00 x 10⁸ m/s) / (510.0 x 10⁻⁹ m)
E = 3.898 x 10⁻¹⁹ J
Finally, we can find the number of photons that strike the area per second by dividing the average power by the energy carried by each photon:
Number of photons = (4000 J·s⁻¹·m⁻²) / (3.898 x 10⁻¹⁹ J)
Number of photons = 1.02 x 10²² photons
Therefore, approximately 1.02 x 10²² photons strike a 6.60 cm² area per second.
E of 510 nm photon = hc/wavelength
Substitute and solve for E in J and I get about 4E-19 J but you need to do that more accurately.
We need to change the energy hitting Madison to cm^2 and not m^2 so
4000 J/sec*m^2 x (1 m/100 cm) x (1 m/100 cm) = about 0.4 J/sec*cm^2 and that x 6.60 cm^2 = about 2.6 J/sec*cm^2. Again that's only an estimate.
Then 4E-19 J/photon x #photons = 2.6 J.
Solve for # photons.