GR9677 #59
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Alternate Solutions |
insertphyspun 2011-05-26 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 2009-09-10 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
 |  |
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Comments |
digitalhikes 2019-07-08 11:04:17 | Happy to go to your post, i\\\\\\\'m by all accounts forward to more and more dependable articles and that i figure we have a tendency to as a full would like to convey such Brobdingnagian numbers of excellent articles, post like Hostpapa Promo Code 2019 to important to us. \\\\r\\\\n |  | insertphyspun 2011-05-26 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 2011-10-07 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.
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|  | proctort 2009-09-10 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 2012-10-02 01:57:42 |
Doesn't the wavelength = 600*10^-9? This gives me an answer of 1/3*10^10...
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calcuttj 2014-09-20 07:21:37 |
Technically the wavelength is 6E-7 or 600E-9
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|  | proctort 2009-09-10 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 2008-10-18 13:17:55 | where does that 1.602E-19 come from?
akbar5223 2008-10-28 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 = 1.602E-19 J.
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naroays 2008-11-03 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
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naroays 2008-11-03 02:51:08 |
Ah, I just noticed the soln in the website doesn't use SI units. Nevermind
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|  |
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