Photon - h/(lamda * c)

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  • #1
Rel
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Main Question or Discussion Point

Hello

I red in web side " The photon has mass " Is that real ?


and what about


m = h / lamda * c
 

Answers and Replies

  • #2
CompuChip
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Hi, welcome to PF.

There is indeed a certain energy assigned to the photon, which is related to its frequency [itex]\omega = 2\pi\nu = 2\pi c / \lambda[/itex] by
[tex]E = \hbar \omega = \frac{h}{2 \pi} \omega = \frac{h c}{\lambda}[/tex]
When we invoke the famous [itex]E = mc^2[/itex], the equivalence between energy and mass, we can indeed think of this energy as representing a mass
[tex]m = \frac{E}{c^2} = \frac{h}{\lambda c}[/tex]

The "rest" mass is usually defined by
[tex](m_0c^2)^2 = E^2 - (\vec p c)^2 [/tex]
where p is the (three)-momentum. If we take this momentum to be [itex]\vec p^2 = m^2 \vec v^2 = (m c)^2 = (h / \lambda)^2[/tex] then this formula gives
[tex](m_0c^2)^2 = (h c / lambda)^2 - (h c / lambda)^2[/tex]
so [itex]m_0 = 0[/itex]. It should, because no massive particle can travel faster than light.

These considerations are important in processes such as pair production and annihilation, where conservation of 'mass' (= energy) must hold.
 
  • #3
Rel
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Thank you compuchip

You gived to me what I want
 

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