If E=hf is applicable for electron and other particles, the De Broglie wavelength should be λ=hv/pc. Because, mc^2=hf which implies mc^2=hc/λ which implies m=h/λc and thus λ=hv/pc. But I have found in my text book that λ=h/p is applicable not only for photon but also for all particle. But how can λ=h/p=h/mv be applicable for all particle?
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Related : About de Broglie relations, what exactly is E ? Its energy of what?. – Frobenius Nov 06 '20 at 13:13
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This video from 3blue1brown has a section on this. The more general uncertainty principle, beyond quantum – mmesser314 Nov 06 '20 at 14:52
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The De Broglie relations $$ E=h\nu, p=\frac{h}{\lambda} $$ are applicable to all particles. However the relation between the frequency/energy and the wave length/momentum, is not the same. Thus, for photons $f=\frac{c}{\lambda}$, for non-relativistic electrons $E=\frac{p^2}{2m}$, for relativistic electrons $E^2=m^2c^4 + p^2c^2$.
Roger V.
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For your last step you just introduce a v in both the numerator and denominator. That is meaningless as it is unity.
In the expression $\lambda = \frac{h}{mc}$, you just have to set $c=v$ to account for particles with different momenta and thus different velocities, not equal to the speed of light.
Do it in this way:
$$E = hf = mc^2 $$ $$\frac{hc}{\lambda} = mc^2$$ $$\frac{h}{\lambda} = mc$$ $$\lambda = \frac{h}{mc}$$ $$\lambda = \frac{h}{mv}$$
Alon Shoshan
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I have found λ=h/mv in some text books. Isn't it incorrect? – Mohammad Hasibul Hossain Rafi Nov 07 '20 at 07:56
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But why and how have you put v in the last line? – Mohammad Hasibul Hossain Rafi Dec 11 '20 at 12:27
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Because not all particles move at the speed of light c, so you can set c=v to account for particles with different moments and so different velocity. – Alon Shoshan Dec 12 '20 at 13:19
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You clearly asked about the De Broglie wavelength for particles other than photons. So I repeat, not all particles move at the speed of light. – Alon Shoshan Dec 12 '20 at 13:21