Doubts
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Re: Doubts
Here's another vote for (b).
If it were an infinite mass atom then it would be (a). With a finite mass, some of the energy of the photon will become kinetic energy of the energized atom. This means you need a little extra energy hence (b).
If it were an infinite mass atom then it would be (a). With a finite mass, some of the energy of the photon will become kinetic energy of the energized atom. This means you need a little extra energy hence (b).
Re: Doubts
To get the answer exactly, you can use the relation $$E^2 = (pc)^2 + (m_0c^2)^2,$$ and the fact that the momentum of the atom after the collision is the same as the momentum of the photon, to get $$(Mc^2 + fh)^2 = (fh)^2 + ((M + \Delta)c^2)^2,$$ and then solve for f.
- WhoaNonstop
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Re: Doubts
Solving for f looks too haaaaaaaaard.kroner wrote:To get the answer exactly, you can use the relation $$E^2 = (pc)^2 + (m_0c^2)^2,$$ and the fact that the momentum of the atom after the collision is the same as the momentum of the photon, to get $$(Mc^2 + fh)^2 = (fh)^2 + ((M + \Delta)c^2)^2,$$ and then solve for f.
-Riley
- HappyQuark
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Re: Doubts
That's why Mathematica has the solve function. Just the other day I came across some really difficult algebra and had to have Mathematica break it down for me. How in the hell am I supposed to know how to solve x+2=4?WhoaNonstop wrote:Solving for f looks too haaaaaaaaard.kroner wrote:To get the answer exactly, you can use the relation $$E^2 = (pc)^2 + (m_0c^2)^2,$$ and the fact that the momentum of the atom after the collision is the same as the momentum of the photon, to get $$(Mc^2 + fh)^2 = (fh)^2 + ((M + \Delta)c^2)^2,$$ and then solve for f.
-Riley
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Re: Doubts
Mind sharing where these problems are from?
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- Joined: Sun Jun 27, 2010 8:15 am
Re: Doubts
Well these problems are from last year's entrance exam for PhD(physics) program at TIFR,india...