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Recoil velocity of $Na$ atom after ejecting a photon.

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A $Na$ atom emits a photon of wavelength $590\ nm$ and recoils with velocity $v$, find $v$?

The solution provided is as :

$v=\dfrac{h}{m\lambda}=\dfrac{6.626\times 10^{-34}}{(23\times 1.67\times 10^{-27})\cdot (590\times 10^{-9})}$.

Now my doubt is that according to De Broglie hypothesis a particle of mass $m$ moving with speed $v$ behaves in some ways like waves of wavelength $\lambda$, but how the author can use $\lambda$ given in problem as it is of the photon not of $Na$

asked Jul 2, 2018 in Physics Problems by n3 (508 points)
edited Jul 2, 2018 by sammy gerbil
Ayush please can you rate my answers to your 3 previous questions. You made some comments to "Problem on Capacitors" but did not state what you think the solution should be. You have not made any comments to the qns on Refraction and Induction. You have not voted for any answers nor selected them as Best Answer. If you think they are not correct please explain.
Not that way, your answers really helped me, Thanks!

1 Answer

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Best answer

In this calculation $\lambda$ refers to the wavelength of the photon, not the de Broglie wavelength of the $Na$ atom.

The momentum of a photon is $p=\frac{E}{c}=\frac{hf}{c}=\frac{h}{\lambda}$. The relation between (de Broglie) wavelength and momentum is the same for material particles as well as massless particles like the photon. The recoil momentum of the $Na$ atom is the same as the momentum of the emitted photon; therefore the de Broglie wavelengths are also the same.

answered Jul 2, 2018 by sammy gerbil (28,448 points)
selected Jul 2, 2018 by n3
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