Answer:
red laser emits 1.42 times the number of photons of the blue laser, per unit of time
Explanation:
In order to calculate the rates of photon emission for both wavelengths, you take into account that the power of the light is given by the following formula:
[tex]P=\frac{E}{t}=\frac{hc/\lambda}{t}[/tex] (1)
That is, the power is the energy per time.
h: Planck's constant
c: speed of light
λ: wavelength of the light
The number of photons emitted per unit of time is given by:
[tex]n=\frac{P}{E}[/tex]
P: power of the light
E: energy of the light
For the two wavelengths you have
[tex]n_1=\frac{P_1}{hc/\lambda_1}\\\\n_2=\frac{P_2}{hc/\lambda_2}\\\\\frac{n_1}{n_2}=\frac{\lambda_1}{\lambda_2}[/tex] (2)
Where you have use that P1=P2
Finally, you replace the values of the wavelengths in the equation (2):
[tex]\frac{n_1}{n_2}=\frac{670nm}{470nm}[/tex][tex]\ = 1.42\frac{photons}{s}[/tex]
Then, the red laser emits 1.42 times the number of photons emited by the blue laser
The rate of photon emission for the blue laser and red laser is [tex]E_b = 1.426 E_r[/tex].
The given parameters;
The energy of a photon emitted by the red laser is calculated as follows;
[tex]E = hf\\\\E = \frac{hc}{\lambda} \\\\E_1 \lambda _1 = E_2 \lambda_2 \\\\\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1} \\\\\frac{E_b}{E_r} = \frac{670 \ nm}{470 \ nm} \\\\\frac{E_b}{E_r} = 1.426\\\\E_b = 1.426 E_r[/tex]
Thus, the rate of photon emission for the blue laser and red laser is [tex]E_b = 1.426 E_r[/tex].
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