# de Broglie questions and answers

Recent questions in de Broglie Equation
Karissa Sosa 2022-05-20 Answered

### Calculating an energy of an electron with known De Broglie wavelength (why can't we calculate it similar than we do it for a photon)Lets say we have an electron with known De Broglie wavelength $\lambda$. Can anyone justify or explain why we calculate its energy $E$ using 1st the De Broglie relation $\lambda =h/p$ to get momentum $p$ and 2nd using the invariant interval to calculate $E$$\begin{array}{rl}{p}^{2}{c}^{2}& ={E}^{2}-{{E}_{0}}^{2}\\ E& =\sqrt{{p}^{2}{c}^{2}+{{E}_{0}}^{2}}\end{array}$Why we are not alowed to do it like we do it for a photon:$\begin{array}{r}E=h\nu =h\frac{c}{\lambda }\end{array}$These equations return different results.

Dennis Montoya 2022-05-18 Answered

### Combination of de Broglie wavelength and mass–energy equivalence gone wrong?I tried to combine the mass–energy equivalence for a particle with mass,$E=\sqrt{\left(m{c}^{2}{\right)}^{2}+\left(pc{\right)}^{2}}=\sqrt{\left(m{c}^{2}{\right)}^{2}+\left(\gamma mvc{\right)}^{2}}$with de Broglie wavelength,$\lambda =\frac{h}{p}=\frac{h}{\gamma mv}.$I get this equation:$E=\frac{h{c}^{2}}{\lambda v}.$This does not seem right, since the equations suggest the energy increases as the speed slow down which is not the case. But I can't see what I did wrong, either. Can someone help me?

Regina Ewing 2022-05-18 Answered

### De Broglie wavelength of an electronIf an electron which already possesses some kinetic energy of $X\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}$ is further accelerated through a potential difference of $X\prime \phantom{\rule{thinmathspace}{0ex}}\mathrm{V}$, then is it correct to state that the electron now has a total kinetic energy of $\left(X+X\prime \right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}$? Using this information, am I allowed to substitute this value in the de Broglie equation to find the wavelength of the electron?

Lexi Chandler 2022-05-15 Answered

### Is the relation c=νλ valid only for Electromagnetic waves?What is the validity of the relation $c=\nu \lambda$? More specifically, is this equation valid only for Electromagnetic waves?I read this statement in a book, which says:"de Broglie waves are not electromagnetic in nature, because they do not arise out of accelerated charged particle."This seems correct, but arises a doubt in my mind.Suppose I find out the wavelength of a matter wave (or de Broglie wave) using de Broglie's wave equation:$\lambda =\frac{h}{p}$Now, can I use $c=\nu \lambda$ to find out the frequency of the wave?

Jace Wright 2022-05-14 Answered

### Derivation of time-dependent Schrödinger equation from De Broglie hypothesisIn a quantum mechanics script I'm reading, the Schrödinger equation is "derived" (more precisely, motivated) by the De Broglie hypothesis. It starts at$\lambda =\frac{2\pi h}{p}$$\omega =\frac{E}{h}$then takes the partial derivatives of the wave $\mathrm{\Psi }\left(x,t\right)=\mathrm{\Psi }\left(0,0\right){e}^{\frac{2\pi ix}{\lambda }-it\omega }$$\begin{array}{}\text{(1)}& \frac{\mathrm{\partial }}{\mathrm{\partial }x}\mathrm{\Psi }\left(x,t\right)=\frac{ip}{h}\mathrm{\Psi }\left(x,t\right)\end{array}$$\frac{\mathrm{\partial }}{\mathrm{\partial }t}\mathrm{\Psi }\left(x,t\right)=-i\frac{E}{h}\mathrm{\Psi }\left(x,t\right)$With the non-relativistic free particle $E=\frac{1}{2m}{p}^{2}$ one gets$\begin{array}{}\text{(2)}& ih\frac{\mathrm{\partial }}{\mathrm{\partial }t}\mathrm{\Psi }\left(x,t\right)=E\mathrm{\Psi }\left(x,t\right)=\frac{{p}^{2}}{2m}\mathrm{\Psi }\left(x,t\right)\end{array}$From there, they miraculously get the time-dependent Schrödinger equation. I cannot understand this step. If I insert formula (1) for ${p}^{2}$ in (2), I get something with $\left(\frac{\mathrm{\partial }}{\mathrm{\partial }x}\mathrm{\Psi }{\right)}^{2}$, which is not the second partial derivative of $\mathrm{\Psi }$ with respect to $x$.Any hints?

Jaeden Weaver 2022-05-14 Answered

### Proof of de Broglie wavelength for electronAccording to de Broglie's wave-particle duality, the relation between electron's wavelength and momentum is $\lambda =h/mv$The proof of this is given in my textbook as follows:1.De Broglie first used Einstein's famous equation relating matter and energy,$E=m{c}^{2},$where $E=$ energy, $m=$ mass, $c=$ speed of light.2.Using Planck's theory which states every quantum of a wave has a discrete amount of energy given by Planck's equation,$E=h\nu ,$where $E=$ energy, $h=$ Plank's constant ($6.62607×{10}^{-34}\phantom{\rule{mediummathspace}{0ex}}\mathrm{J}\phantom{\rule{mediummathspace}{0ex}}\mathrm{s}$), $\nu =$ frequency.3.Since de Broglie believes particles and wave have the same traits, the two energies would be the same:$m{c}^{2}=h\nu .$$m{c}^{2}=h\nu .$4.Because real particles do not travel at the speed of light, de Broglie substituted $v$, velocity, for $c$, the speed of light:$m{v}^{2}=h\nu .$I want a direct proof without substituting $v$ for $c$. Is it possible to prove directly $\lambda =h/mv$ without substituting $v$ for $c$ in the equation $\lambda =h/mc$?

Jaylene Duarte 2022-05-14 Answered