Doubts

 Posts: 381
 Joined: Mon May 24, 2010 11:34 pm
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
 Posts: 853
 Joined: Mon Sep 21, 2009 1:31 am
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
 Posts: 762
 Joined: Thu Apr 16, 2009 2:08 am
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

 Posts: 4
 Joined: Sun Jun 27, 2010 8:15 am
Re: Doubts
Mind sharing where these problems are from?

 Posts: 4
 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...