Assuming that 9.0\% of the energy output of a 65W lightbulb is visible light and that the average wavelength of the light is 550 {\rm nm}, how many photons are emitted by the lightbulb each second?
Number photons*energy each=.09(65)
energy each=hf=h*c/lambda
To determine the number of photons emitted by the lightbulb each second, we can use the following steps:
Step 1: Calculate the energy of the lightbulb using the Power-Energy relationship.
The energy emitted by the lightbulb can be calculated using the formula:
Energy = Power × time
Given that the power output of the lightbulb is 65W, we need to determine the energy output over a time interval of 1 second.
Energy = 65W × 1s = 65 J
Step 2: Calculate the energy of the visible light using the given percentage.
The percentage of energy output that corresponds to visible light is given as 9.0%. To find the energy of the visible light, we can convert the given percentage to a decimal.
Energy of visible light = (9.0/100) × total energy
= (9.0/100) × 65 J
= 5.85 J
Step 3: Calculate the number of photons using the energy of the visible light and the energy of a single photon.
The energy of a single photon can be calculated using Planck's equation:
E = hf
where E is the energy of a single photon, h is Planck's constant (6.63 × 10^(-34) J·s), and f is the frequency of the light.
To find the energy of a single photon, we need to calculate the frequency of the light using its wavelength.
The relationship between wavelength (λ), frequency (f), and the speed of light (c) is given by:
c = λf
Rearranging the equation, we get:
f = c/λ
Given that the average wavelength of the light is 550 nm (or 550 × 10^(-9) m) and the speed of light is approximately 3 × 10^8 m/s, we can calculate the frequency:
f = (3 × 10^8 m/s) / (550 × 10^(-9) m)
= 5.45 × 10^14 Hz
Now, we can calculate the energy of a single photon:
E = (6.63 × 10^(-34) J·s) × (5.45 × 10^14 Hz)
= 3.60 × 10^(-19) J
Finally, we can find the number of photons by dividing the energy of the visible light by the energy of a single photon:
Number of photons = Energy of visible light / Energy of a single photon
= 5.85 J / (3.60 × 10^(-19) J)
≈ 1.63 × 10^19 photons
Therefore, approximately 1.63 × 10^19 photons are emitted by the lightbulb each second.