GREPhysics.NET: GR9677 #59

GR9677 #59

Problem

GREPhysics.NET Official Solution

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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

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. Reply to this comment
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 $n = \frac{10^4 \cdot 6 \times 10^{-9} \cdot 10^{-15}}{6.6 \times 10^{-34} \cdot 3 \times 10^8} \approx \frac{1}{3}\times 10^8 \approx 10^7$ Reply to this comment

Comments

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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.

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 $E = n \frac{h c}{\lambda}$ , 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
$n = \frac{10^4 \cdot 6 \times 10^{-9} \cdot 10^{-15}}{6.6 \times 10^{-34} \cdot 3 \times 10^8} \approx \frac{1}{3}\times 10^8 \approx 10^7$

alisonsparkles
2012-10-02 01:57:42

Doesn't the wavelength = 600*10^-9? This gives me an answer of 1/3*10^10...

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 $E = n \frac{h c}{\lambda}$ , 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
$n = \frac{10^4 \cdot 6 \times 10^{-9} \cdot 10^{-15}}{6.6 \times 10^{-34} \cdot 3 \times 10^8} \approx \frac{1}{3}\times 10^8 \approx 10^7$

Reply to this comment

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.

naroays
2008-11-03 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

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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You are replying to:

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 $n = \frac{10^4 \cdot 6 \times 10^{-9} \cdot 10^{-15}}{6.6 \times 10^{-34} \cdot 3 \times 10^8} \approx \frac{1}{3}\times 10^8 \approx 10^7$

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