MHB How Many Photons Does a Laser Emit in a Single Pulse?

[yakin]

A laser emits light of wavelength 463 nm during a brief pulse that lasts for 25 ms and has a total energy of 1.2 J. How many photons are emitted in that single pulse? (c=3.00E8 m/s, h=6.626E-34 j.s)

 

[DavidCampen]

You asked essentially the same question here less than 2 weeks ago.
http://mathhelpboards.com/other-topics-22/physics-84-kw-am-radio-station-broadcasts-1000-khz-how-many-photons-emitted-each-second-10382.html?highlight=radio

 

[SuperSonic4]

A laser emits light of wavelength 463 nm during a brief pulse that lasts for 25 ms and has a total energy of 1.2 J. How many photons are emitted in that single pulse? (c=3.00E8 m/s, h=6.626E-34 j.s)

The trick here is knowing that the energy of an individual photon is given by $$E = h\nu = \dfrac{hc}{\lambda}$$

You can then divide 1.2 by your answer to get the number of photons. The 25ms (unless I'm missing something) is a red herring

 

[sweer6]

Energy of a photon of light
E = hf = hc/l
where
E = energy of radiation (J)
h = Planck's constant = 6.626*10^(-34) J.s.
f = frequency of radiation (Hz)
c = speed of light = 3.8 m/s
l = (lambda) wavelength of radiation (m)

You need to convert all your data to the units above or at least have the same units all the way through. Such as f = 1/t or the inverse of period. The period is given as 25 ms which you need to convert to metres first and then take the inverse of, to get the frequency.
Hope this helps

 

[zzephod]

Energy of a photon of light
E = hf = hc/l
where
E = energy of radiation (J)
h = Planck's constant = 6.626*10^(-34) J.s.
f = frequency of radiation (Hz)
c = speed of light = 3.8 m/s
l = (lambda) wavelength of radiation (m)

You need to convert all your data to the units above or at least have the same units all the way through. Such as f = 1/t or the inverse of period. The period is given as 25 ms which you need to convert to metres first and then take the inverse of, to get the frequency.
Hope this helps

The wavelength $$\lambda$$ is given explicitly in the question as 463nm, so the period is not 25 ms which is the duration of the pulse and irrelevant to the question as asked.

.

 

[sweer6]

The wavelength $$\lambda$$ is given explicitly in the question as 463nm, so the period is not 25 ms which is the duration of the pulse and irrelevant to the quation as asked.

.

ok, let me get this straight first, what does 25 ms mean? What unit is it measured in?

 

[zzephod]

ok, let me get this straight first, what does 25 ms mean? What unit is it measured in?

milli-seconds

 

[I like Serena]

c = speed of light = 3.8 m/s

Hmm, a photon has the same speed as a bicyclist?

 

[sweer6]

Hmm, a photon has the same speed as a bicyclist?

yeah, yeah, i made a mistake, lol , the speed of light is 3*10^8 m/s

 

[Deveno]

Hmm, a photon has the same speed as a bicyclist?

Well, if that bicyclist is "The Flash", sure.

 

[chisigma]

A laser emits light of wavelength 463 nm during a brief pulse that lasts for 25 ms and has a total energy of 1.2 J. How many photons are emitted in that single pulse? (c=3.00E8 m/s, h=6.626E-34 j.s)

The question is interesting because it reminds me of when I worked on digital transmission in optical fiber. If an optical pulse of power E is composed of N photons each of wavelength $\displaystyle \lambda_{i}= \frac{c}{\nu_{i}}$ then is...

$\displaystyle E = h\ \sum_{i=1}^{N} \frac{c}{\lambda_{i}}\ (1)$

We have to use (1) because in the famous Plank's expression $\displaystyle \varepsilon = h\ \nu$ h is a constant but $\nu$ is a continouos random variable and the same is for $\lambda$. In other words that means that in nature there are no two-photon equal and if You in (1) know the value of E, h, c, and the mean value $\displaystyle \overline \lambda$ of the photons, the N must be considered as a random variable and what You can obtain is its mean value $\overline N$ which is not necessarly an integer. Thi non coventional approach allowed me to solve in very comfortable way the problem of a time spreading of an optical pulse along an optical fibre due to chromatic dispersion...

Kind regards

$\chi$ $\sigma$