To determine how many photons are emitted per second by a heat lamp producing 27.0 watts of power at a wavelength of 6.0 nanometers (which is [tex]\( 6.0 \times 10^{-7} \)[/tex] meters), we can follow these steps:
1. Understand the relationships and constants:
- Power (P) = [tex]\( 27.0 \)[/tex] watts (1 watt equals 1 Joule per second).
- Wavelength ([tex]\(\lambda\)[/tex]) = [tex]\( 6.0 \times 10^{-7} \)[/tex] meters.
- Speed of light (c) = [tex]\( 3.0 \times 10^{8} \)[/tex] meters per second.
- Planck's constant (h) = [tex]\( 6.626 \times 10^{-34} \)[/tex] Joule seconds.
2. Calculate the energy of a single photon (E):
The energy of a photon is given by the equation [tex]\( E = \frac{hc}{\lambda} \)[/tex].
- [tex]\( h = 6.626 \times 10^{-34} \)[/tex] Joule seconds.
- [tex]\( c = 3.0 \times 10^{8} \)[/tex] meters per second.
- [tex]\( \lambda = 6.0 \times 10^{-7} \)[/tex] meters.
Plugging in these values, we get:
[tex]\[
E = \frac{(6.626 \times 10^{-34} \, \text{J s}) (3.0 \times 10^{8} \, \text{m/s})}{6.0 \times 10^{-7} \, \text{m}}
\][/tex]
Simplifying this, we get the energy per photon.
3. Determine the number of photons emitted per second (N):
The number of photons emitted per second is given by [tex]\( N = \frac{P}{E} \)[/tex].
- Power [tex]\( P = 27.0 \)[/tex] watts (27 Joules per second).
Using the energy per photon [tex]\( E \)[/tex] calculated in the previous step, we can determine [tex]\( N \)[/tex].
4. Express the answer:
After performing the calculation, we find that the number of photons emitted per second is:
[tex]\[
N = 8.15 \times 10^{19} \, \text{photons/second}
\][/tex]
(rounded to three significant figures).
Therefore, the heat lamp emits approximately [tex]\( 8.15 \times 10^{19} \)[/tex] photons per second.
