GR9677 #59



Alternate Solutions 
insertphyspun 20110526 13:46:20  Of course, the E=Pt approach is probably best, but I was going too fast to see that P=10kW. My alternate solution uses the uncertainty principle:
where . Thus,
Close enough, and no need to convert Watts to eV/s.   proctort 20090910 20:00:28  First time posting, defaults to NEC apparently. This one is properly labeled:
Given that this answer is virtually illegible, I'm writing out my own answer:
The energy of a beam of light is , where is the number of photons and is the wavelength. As power is defined as P=\frac{E}{t}, where is the time over which energy is given off, we have , which, solving for gives n = \frac{P \lambda t}{h c}.
In SI units, , ..., which gives
 

Comments 
insertphyspun 20110526 13:46:20  Of course, the E=Pt approach is probably best, but I was going too fast to see that P=10kW. My alternate solution uses the uncertainty principle:
where . Thus,
Close enough, and no need to convert Watts to eV/s.
Setareh 20111007 09:52:25 
I think in this case you have forgotten "c" in denominator of n=lambda/4*pi*t. rnActually the real equation for n is:rnn=lambda/4*pi*c*trnif you put c= 3*10^8 , you will see that it is larger than the correct answer.rnI think you have to write: n<=lambda/4*pi*c*trnyou have only found the boundary, but the power will determine which n is eligible for this problem.

  proctort 20090910 20:00:28  First time posting, defaults to NEC apparently. This one is properly labeled:
Given that this answer is virtually illegible, I'm writing out my own answer:
The energy of a beam of light is , where is the number of photons and is the wavelength. As power is defined as P=\frac{E}{t}, where is the time over which energy is given off, we have , which, solving for gives n = \frac{P \lambda t}{h c}.
In SI units, , ..., which gives
alisonsparkles 20121002 01:57:42 
Doesn't the wavelength = 600*10^9? This gives me an answer of 1/3*10^10...

calcuttj 20140920 07:21:37 
Technically the wavelength is 6E7 or 600E9

  proctort 20090910 19:53:33  Given that this answer is virtually illegible, I'm writing out my own answer:
The energy of a beam of light is , where is the number of photons and is the wavelength. As power is defined as , where is the time over which energy is given off, we have , which, solving for gives .
In SI units, , ..., which gives
  isina 20081018 13:17:55  where does that 1.602E19 come from?
akbar5223 20081028 08:12:41 
The laser power is given in W=J/s, so it is necessary to convert the power from J/s to eV/s, where 1 eV = 1.602E19 J.

naroays 20081103 02:41:15 
Why should we convert to eV/s?
The SI unit of energy is Joules, and is in Joules, as is Power*time, because it's Joules/second * second = Joules

naroays 20081103 02:51:08 
Ah, I just noticed the soln in the website doesn't use SI units. Nevermind

 

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