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Sagot :
To determine the number of photons emitted by a bulb of 40 W producing light of wavelength 620 nm with [tex]\(80 \%\)[/tex] efficiency over a period of 20 seconds, follow these steps:
1. Calculate the energy produced by the bulb in 20 seconds:
The power (P) of the bulb is 40 W, and the time (t) period is 20 seconds. Energy (E) produced is given by:
[tex]\[ E = P \times t \][/tex]
Substituting the given values:
[tex]\[ E = 40 \, \text{W} \times 20 \, \text{s} = 800 \, \text{J} \][/tex]
2. Calculate the energy of one photon:
The energy (E_photon) of a single photon can be calculated using the formula:
[tex]\[ E_{\text{photon}} = \frac{h \cdot c}{\lambda} \][/tex]
where:
- [tex]\( h \)[/tex] is the Planck constant ([tex]\(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\)[/tex])
- [tex]\( c \)[/tex] is the speed of light ([tex]\(3 \times 10^8 \, \text{m/s}\)[/tex])
- [tex]\( \lambda \)[/tex] is the wavelength ([tex]\(620 \times 10^{-9} \, \text{m}\)[/tex])
Plugging in the values:
[tex]\[ E_{\text{photon}} = \frac{6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \times 3 \times 10^8 \, \text{m/s}}{620 \times 10^{-9} \, \text{m}} \approx 3.206 \times 10^{-19} \, \text{J} \][/tex]
3. Calculate the actual energy used to produce light:
Given the efficiency (η) is 80%, the actual energy used to produce light (E_luminous) is:
[tex]\[ E_{\text{luminous}} = E \times \eta \][/tex]
Substituting the given efficiency:
[tex]\[ E_{\text{luminous}} = 800 \, \text{J} \times 0.80 = 640 \, \text{J} \][/tex]
4. Calculate the number of photons emitted:
The number of photons [tex]\(N\)[/tex] emitted can be obtained by dividing the luminous energy by the energy per photon:
[tex]\[ N = \frac{E_{\text{luminous}}}{E_{\text{photon}}} \][/tex]
Substituting the values:
[tex]\[ N = \frac{640 \, \text{J}}{3.206 \times 10^{-19} \, \text{J/photon}} \approx 1.996 \times 10^{21} \][/tex]
Therefore, the option closest to the calculated number of photons emitted is:
[tex]\[ \boxed{2 \times 10^{21}} \][/tex]
1. Calculate the energy produced by the bulb in 20 seconds:
The power (P) of the bulb is 40 W, and the time (t) period is 20 seconds. Energy (E) produced is given by:
[tex]\[ E = P \times t \][/tex]
Substituting the given values:
[tex]\[ E = 40 \, \text{W} \times 20 \, \text{s} = 800 \, \text{J} \][/tex]
2. Calculate the energy of one photon:
The energy (E_photon) of a single photon can be calculated using the formula:
[tex]\[ E_{\text{photon}} = \frac{h \cdot c}{\lambda} \][/tex]
where:
- [tex]\( h \)[/tex] is the Planck constant ([tex]\(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\)[/tex])
- [tex]\( c \)[/tex] is the speed of light ([tex]\(3 \times 10^8 \, \text{m/s}\)[/tex])
- [tex]\( \lambda \)[/tex] is the wavelength ([tex]\(620 \times 10^{-9} \, \text{m}\)[/tex])
Plugging in the values:
[tex]\[ E_{\text{photon}} = \frac{6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \times 3 \times 10^8 \, \text{m/s}}{620 \times 10^{-9} \, \text{m}} \approx 3.206 \times 10^{-19} \, \text{J} \][/tex]
3. Calculate the actual energy used to produce light:
Given the efficiency (η) is 80%, the actual energy used to produce light (E_luminous) is:
[tex]\[ E_{\text{luminous}} = E \times \eta \][/tex]
Substituting the given efficiency:
[tex]\[ E_{\text{luminous}} = 800 \, \text{J} \times 0.80 = 640 \, \text{J} \][/tex]
4. Calculate the number of photons emitted:
The number of photons [tex]\(N\)[/tex] emitted can be obtained by dividing the luminous energy by the energy per photon:
[tex]\[ N = \frac{E_{\text{luminous}}}{E_{\text{photon}}} \][/tex]
Substituting the values:
[tex]\[ N = \frac{640 \, \text{J}}{3.206 \times 10^{-19} \, \text{J/photon}} \approx 1.996 \times 10^{21} \][/tex]
Therefore, the option closest to the calculated number of photons emitted is:
[tex]\[ \boxed{2 \times 10^{21}} \][/tex]
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