De Broglie wavelength Vs Wavefunction - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-09-20T23:15:02Z https://physics.stackexchange.com/feeds/question/188216 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/188216 1 De Broglie wavelength Vs Wavefunction Shahbaaz1104 https://physics.stackexchange.com/users/83081 2015-06-07T19:13:06Z 2015-06-09T15:40:31Z <p>According to De Broglie's eqn by calculating momentum of electron in 1st orbit of hydrogen, the wavelength is itself the diameter of the atom, so what does it tell abt the wave function? In general what is the relationship between De Broglie's eqn and wave function? In case my question is wrong , then please help me correct my thinking. I am a 11th std student.</p> https://physics.stackexchange.com/questions/188216/-/188570#188570 1 Answer by Ian for De Broglie wavelength Vs Wavefunction Ian https://physics.stackexchange.com/users/76310 2015-06-09T15:40:31Z 2015-06-09T15:40:31Z <p>The wave function in an atom can be written as a sum of wave functions of the form</p> <p>$\Psi_{\vec{k}}(\vec{r})=e^{i\vec{k}\cdot\vec{r}}$</p> <p>In using deBroglie's equation $p=\hbar k$ to estimate the momentum of an electron in a hydrogen atom, you are assuming that the dominant wave function $\Psi_{\vec{k_0}}$ in the sum has a wave vector $\vec{k_0}$ such that</p> <p>$$k_0=\frac{2\pi}{\lambda_0}$$</p> <p>where $\lambda_0$ is the diameter of the atom. In this way, you can see how using deBroglie's equation to estimate the average momentum of an electron in an atom is actually a crude approximation that leaves out all the terms in the wave function with momenta different from $\hbar k_0$.</p>