Find number of photons emmitted per sec?

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Homework Help Overview

The discussion revolves around calculating the number of photons emitted per second by a Gallium Arsenide diode laser, which emits 0.9 mW of power. The bandgap energy of the material is given as 1.42 eV, and participants are exploring how to use this information to find the desired quantity.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between power, energy, and the number of photons, questioning how to combine the energy emitted per second with the energy per photon to find the number of emitted photons. There are inquiries about the relevance of the bandgap and the absence of wavelength information.

Discussion Status

Several participants have provided calculations and attempted to clarify the relationships between the quantities involved. There is a mix of confirmations and corrections regarding the calculations, with some participants expressing uncertainty about unit conversions and the final result.

Contextual Notes

Some participants note the need for unit conversions and question the assumptions made regarding the energy of the emitted photons. There is also a mention of potential errors in numerical calculations.

Kobayashi
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Hi, I have a question asking me to find the number of photons emmitted per second. I am given the information that the Gallium Arsenide diode laser emit 0.9 mW of power inside a CD player. The bandgap of the Gallium Arsenide is 1.42 eV. What equation is used to find this?
 
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P = E/t (power, energy, time)
E= n (1 photon energy)

n: piece
 
How come you're not given the wavelength of the emitted photons?

EDIT: Ok, I assume it's because the energy of each photon emitted would be equivalent to something which is already given in the question.
 
Last edited:
Where is the bandgap taken into account?
 
I need the bandwidth
 
Two pieces of information are needed to solve this:

1. The amount of energy emitted per sec.
2. The amount of energy per emitted photon.

Units aside, this information is given directly in the problem statement. The problem is to figure out how to combine the 2 numbers to get the number of emitted photons per second.

Again, how to combine
(energy) / (sec)
and
(energy) / (photon)

in order to get

(photons) / (sec)

p.s. some unit conversions are required here.
 
Okay so I get:

0.0009 Joules/second

and

(1.42 eV) * 1.602 x10^-19 = 2.27 x 10^-19 Joules - Is this correct? Is this the energy per photon?

Does (energy/photon)/(energy/second) not give photons/second.

Is that how the problem is solved?
 
Kobayashi said:
Okay so I get:

0.0009 Joules/second

and

(1.42 eV) * 1.602 x10^-19 = 2.27 x 10^-19 Joules - Is this correct? Is this the energy per photon?
Yes.

Does (energy/photon)/(energy/second) not give photons/second.
No, but you are close.

You wrote:

<br /> \frac{(\frac{\mbox{energy}}{\mbox{photon}})}{(\frac{\mbox{energy}}{\mbox{second}})}<br />

How do the units work out in what you wrote?
 
Hi, thanks for your help.
The only alternative I can think of is the Joules per second divided by the Joules per photon. This gives 3.96 x 10^16 Photons per sec. Is this correct?
 
  • #10
You have the right idea, but entered the numbers into your calculator incorrectly.
 
  • #11
Oh yes, 3.96 x 10^15. Is this the correct final answer. Thanks again for all your help.
 
  • #12
Yes.

You're welcome :smile:
 

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