# Why replace c with v in Broglie hypothesis

I am in XI and I am currently studying atomic orbitals, I stumbled up and am stuck at the following step of de Broglie's hypothesis.

I did a little search on-line and found almost same derivation everywhere, the only place I found a more complex and supposedly accurate formula was the Wikipedia itself, though it did not explain the derivation.

Few such sites:

In the derivation of the formula $$\frac{1}{\lambda}=\frac{h}{m\color{green}c}=\frac{h}{m\color{red}v}$$

we replace $c$ with $v$. I want to ask why?

As of Wikipedia it gives a better formula with the Lorentz factor $\gamma$ :-

$$\frac{1}{\lambda}=\frac{h}{\color{red}\gamma m_oc}=\frac{h}{m_ov} \sqrt{\color{red}1\color{red}-\frac{\color{red}v^\color{red}2}{\color{red}c^\color{red}2}}$$

Can anyone give the derivation or a hint or a logic that helps understand why the derivation from $E=mc^2$ and $E=h\nu$ gives the De Broglie equation.

C is for radioactive rays while v is for the matter waves..here c is for a particle travelling with light speed, i.e., radioactive rays, so we replace it with v velocity of any particle for its matter wave

$E=mc^2$ and $E=hv$. Therefore, $$mc^2=hv$$ Or $$mc^2=\frac{hc}{k}$$, where k=wavelength $$mc=\frac{h}{k}$$ Or, $$k=\frac{h}{mc}$$ Replacing c by v we get $$k=\frac{h}{mv}$$ Note the difference between matter waves and electromagnetic waves. E.W. are associated with electrical and magnetic field while matter waves are not so E.w. travel at 3×10^2 m/s. While matter waves travel with different velocity. The equation of de broglie with c is valid, it can be treated for finding the wavelength of electromagnetic waves, or electrons. But the velocity of matter waves are quite smaller as compared to light speed. So he used a general v in place of c V can be equal to c when the particle is travelling at light speed

• but c is the constant in the eqation $E=mc^2$ then how and why we replace it – Aditya ultra Jun 29 '15 at 18:35
• We consider c as the velocity of the particle travelling in the wave, but a particle can travel at light speed only when it is electromagnetic wave. But for matter wave the velocity is variable. So just for those particles the velocity is v. We assume the velocity if particles in electromagnetic waves to be c so we write that equation – Aneek Jun 29 '15 at 18:39
• can we really replace a constant with a variable of course the speed of light is a property at which it goes and hence is confused with velocity. According to special relativity, c is the maximum speed at which all matter and information in the universe can travel. It is much like Planck'csconstant so I cannot and should not replace the Planck's constant with a variable of same units, or things can go serious wrong. – Aditya ultra Jun 29 '15 at 18:50
• @Adityaultra c is not the maximum speed at which matter can travel. It would require infinite energy to do so. physics.stackexchange.com/questions/1557/… – Cohen_the_Librarian Aug 26 '15 at 19:47

de Broglie suggested that a moving body behaves in certain ways as though it has a wave nature. His conjecture of dual nature of matter was based on two points : $(1).$ dual nature of radiation, and $(2).$ nature loves symmetry.

de Broglie deduced the connection between particle and wave properties from the Einstein-Planck expression for the energy of the electromagnetic wave and the classical result for the momentum of such a wave. The two expressions are,

$E=hf$

and $P=\frac{E}{c}$

From these equations we get,

$\lambda=\frac{h}{P}$

de Broglie said that this equation is a completely general one that is applied to a material particle as well as photons. The waves associated with the moving particle are called matter waves or de Broglie waves.

de Broglie wavelength is given by,

$\lambda=\frac{h}{mv}$

• That hypothesis worked & established as a theory later? – Mockingbird Feb 26 '17 at 11:18
• Yes, the concept of matter waves was used to justify quantization of angular momentum as proposed in Bohr's model. If we consider standing electron wave, then to maintain a standing wave over the circumference of a circular orbit, the wavelength must be an integral fraction of that circumference- 2πr=nλ=nh/mv ,which gives mvr=nh/2π – Mitchell Feb 26 '17 at 11:42