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GR9677 #59
Problem
 GREPhysics.NET Official Solution Alternate Solutions
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Quantum Mechanics$\Rightarrow$}Laser

Not much understanding of lasers is required to solve this one; the basic idea of the problem tests the relation between photon energy and energy from the laser. Recalling the equation $E=hf=hc/\lambda\approx 12E-7/600E-9=2 eV$, equates that to the energy (in eV) calculated from $Pt = 10E3 \times 1E-15/1.602E-19$. The answer comes out to choice (B).

Alternate Solutions
 insertphyspun2011-05-26 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: $\Delta E \Delta t = h/4\pi$ where $E=nhc/\lambda$. Thus, $n = \lambda/4\pi \Delta t = 5 \times 10^7$ Close enough, and no need to convert Watts to eV/s.Reply to this comment proctort2009-09-10 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 $E = n \frac{h c}{\lambda}$, where $n$ is the number of photons and $\lambda$ is the wavelength. As power is defined as P=\frac{E}{t}, where $t$ is the time over which energy is given off, we have $P=n \frac{h c }{\lambda t}$, which, solving for $n$ gives n = \frac{P \lambda t}{h c}. In SI units, $P = 10^4 \mathrm{W}$, $\lambda = 6 \times 10^{-9} \mathrm{m}$..., which gives $n = \frac{10^4 \cdot 6 \times 10^{-9} \cdot 10^{-15}}{6.6 \times 10^{-34} \cdot 3 \times 10^8} \approx \frac{1}{3}\times 10^8 \approx 10^7$Reply to this comment
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:

$\Delta E \Delta t = h/4\pi$

where $E=nhc/\lambda$. Thus,

$n = \lambda/4\pi \Delta t = 5 \times 10^7$

Close enough, and no need to convert Watts to eV/s.
 Setareh2011-10-07 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.
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 $E = n \frac{h c}{\lambda}$, where $n$ is the number of photons and $\lambda$ is the wavelength. As power is defined as P=\frac{E}{t}, where $t$ is the time over which energy is given off, we have $P=n \frac{h c }{\lambda t}$, which, solving for $n$ gives n = \frac{P \lambda t}{h c}.

In SI units, $P = 10^4 \mathrm{W}$, $\lambda = 6 \times 10^{-9} \mathrm{m}$..., which gives
$n = \frac{10^4 \cdot 6 \times 10^{-9} \cdot 10^{-15}}{6.6 \times 10^{-34} \cdot 3 \times 10^8} \approx \frac{1}{3}\times 10^8 \approx 10^7$
 alisonsparkles2012-10-02 01:57:42 Doesn't the wavelength = 600*10^-9? This gives me an answer of 1/3*10^10...
 calcuttj2014-09-20 07:21:37 Technically the wavelength is 6E-7 or 600E-9
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 $E = n \frac{h c}{\lambda}$, where $n$ is the number of photons and $\lambda$ is the wavelength. As power is defined as $P=\frac{E}{t}$, where $t$ is the time over which energy is given off, we have $P=n \frac{h c }{\lambda t}$, which, solving for $n$ gives $n = \frac{P \lambda t}{h c}$.

In SI units, $P = 10^4 \mathrm{W}$, $\lambda = 6 ^{-9} \mathrm{m}$..., which gives
$n = \frac{10^4 \cdot 6 \times 10^{-9} \cdot 10^{-15}}{6.6 \times 10^{-34} \cdot 3 \times 10^8} \approx \frac{1}{3}\times 10^8 \approx 10^7$
isina
2008-10-18 13:17:55
where does that 1.602E-19 come from?
 akbar52232008-10-28 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.
 naroays2008-11-03 02:41:15 Why should we convert to eV/s? The SI unit of energy is Joules, and $\frac{\hbar c}{\lambda}$ is in Joules, as is Power*time, because it's Joules/second * second = Joules
 naroays2008-11-03 02:51:08 Ah, I just noticed the soln in the website doesn't use SI units. Nevermind

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