To find the wavelength of the photons emitted by a 1-watt bulb that emits \(2.0 \times 10^{16}\) photons in 1 second, we can follow these steps:
### Step 1: Understand the relationship between power, energy, and photons
Power (P) is defined as the energy emitted per unit time. The energy of a single photon can be expressed using the formula:
\[
E = \frac{hc}{\lambda}
\]
where:
- \(E\) is the energy of a single photon,
- \(h\) is Planck's constant (\(6.62 \times 10^{-34} \, \text{Js}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{ms}^{-1}\)),
- \(\lambda\) is the wavelength of the photon.
### Step 2: Calculate the total energy emitted in 1 second
Since the power of the bulb is 1 watt, the total energy emitted in 1 second is:
\[
E_{\text{total}} = P \times t = 1 \, \text{W} \times 1 \, \text{s} = 1 \, \text{J}
\]
### Step 3: Relate total energy to the number of photons
The total energy emitted can also be expressed in terms of the number of photons (\(n\)) and the energy of a single photon:
\[
E_{\text{total}} = n \times E
\]
Given that \(n = 2.0 \times 10^{16}\) photons, we can write:
\[
1 \, \text{J} = n \times \frac{hc}{\lambda}
\]
### Step 4: Rearrange the equation to find the wavelength
From the above equation, we can rearrange it to solve for \(\lambda\):
\[
\lambda = \frac{n \times hc}{E_{\text{total}}}
\]
### Step 5: Substitute the known values
Substituting the known values into the equation:
- \(n = 2.0 \times 10^{16}\)
- \(h = 6.62 \times 10^{-34} \, \text{Js}\)
- \(c = 3 \times 10^8 \, \text{ms}^{-1}\)
- \(E_{\text{total}} = 1 \, \text{J}\)
We get:
\[
\lambda = \frac{(2.0 \times 10^{16}) \times (6.62 \times 10^{-34}) \times (3 \times 10^8)}{1}
\]
### Step 6: Calculate the value
Now, let's calculate:
\[
\lambda = (2.0 \times 10^{16}) \times (6.62 \times 10^{-34}) \times (3 \times 10^8)
\]
Calculating this step-by-step:
1. Calculate \(6.62 \times 3 = 19.86\)
2. Then, \(2.0 \times 19.86 = 39.72\)
3. Finally, \(39.72 \times 10^{-34} \times 10^{16} = 39.72 \times 10^{-18} = 3.972 \times 10^{-9} \, \text{m}\)
Thus, we have:
\[
\lambda \approx 3.97 \times 10^{-9} \, \text{m}
\]
### Step 7: Convert to nanometers
To convert meters to nanometers:
\[
3.97 \times 10^{-9} \, \text{m} = 3.97 \, \text{nm} \approx 4 \, \text{nm}
\]
### Conclusion
The wavelength of the photons emitted by the bulb is approximately \(4 \, \text{nm}\).
