The De Broglie Wavelength is the very evidence of dual nature of particles ( specially photons). But if we use the equation lamda =h/mv, and then put the values of the rest mass of photon (=0) or the mass in motion (=1/0) , the equation becomes invalid. Explain
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Welcome to this site! Please consider that this is not a site, where people do your home work for you for free. In order to provide some focused help, you should show some efforts, e.g. tries how to solve the problem. Then it's easier to actually help you understand the concepts. – engineer Sep 24 '19 at 06:47
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2@engineer It doesn't really look like a homework question? – user8736288 Sep 24 '19 at 06:57
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2The "explain" at the end to me looks like a clear sign for an exam question or something similar. But of course, I might be wrong there. – engineer Sep 24 '19 at 07:03
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It is policy here not to answer exam questions. I will downvote any such answer. – my2cts Sep 24 '19 at 07:05
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2I have to say this doesn't look to me like a homework or exam question. – John Rennie Sep 24 '19 at 07:33
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I think you should study, special relativity before asking the question, these questions are not base on Newton theories. – आर्यभट्ट Sep 24 '19 at 07:33
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1...I agree with @John Rennie. – Frobenius Sep 24 '19 at 10:04
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...may be we must admit that not only massive objects but also moving packets of energy have momentum too ??? – Frobenius Sep 24 '19 at 10:16
2 Answers
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The de Broglie wavelength is not given by $\lambda = \frac{h}{mv}$. It is given by $\lambda = \frac{h}{p}$, where $p$ is the momentum of the wave-particle.
The momentum of light isn't zero (even though it's massless), so it has a finite de Broglie wavelength.
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Ok. We know that light can be of different wavelengths. Does it mean that all those different wavelengths have different momentum? – Jdeep Sep 24 '19 at 13:55
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And does it mean that we can find the momentum of light by dividing planks constant by its wavelength (p=h/lambda)? – Jdeep Sep 24 '19 at 14:00
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The correct formula is lambda=h/p. For a photon p=gamma.m.v. your forgot gamma factor. P is not null for a photon even if mass is null due to the gamma factor. In maths this gives 0 times infinity. You can t say that this give zero
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A lot of the confusion on this topic seems to arise from people assuming that p=mγvp=mγv should be the definition of momentum. It really isn't an appropriate definition of momentum, because in the case of m=0m=0 and v=cv=c, it gives an indeterminate form. The indeterminate form can, however, be evaluated as a limit in which mm approaches 0 and E=mγc2E=mγc2 is held fixed. The result is again p=E/cp=E/c. – आर्यभट्ट Sep 24 '19 at 07:20