The question is only asking for energies less than V0.
So for region 0<x<L we have the following
k=\frac{\sqrt(2m(V0-E))}{\hbar}
For region x>L we will have V(x)=V0, and the following:
\frac{d^2\psi}{dx^2}+\frac{2m}{\hbar^2}(E-V0)\psi=0
Is it going to be same k for this region as the...
From the Schrodinger equation, if V(x)=0, we can find the following:
\frac{d^2\psi(x)}{dx^2} =-\frac{2mE}{\hbar^2}\psi=-k^2\psi
I do not know how I am suppose to solve it in terms of wave numbers.
For the well shown below (Attachment or Link belew) solve the time independent one dimensional Schrodinger equation for energies less than V0. You may (and should) leave your answer in terms of a single transcendental equation for the allowed wave numbers.
\frac{-\hbar^2}{2m} ...
I found \Delta x \Delta p=\hbar/\sqrt {2}
which means \Delta x \Delta p is independent from the value of alpha, what do you think? somehow I think I should have gotten \Delta x \Delta p=\hbar/2
Question: A light-emitting diode (LED) made of the semiconductor GaAsP gives off red light \lambda =650nm. what is the band gap for this semiconductor?
I know the E=hc/ \lambda
so it means the band gap is 1240/650=1.9 ev ??
Uncertainty - Harmonic Oscillator
The Wave function for the ground state of a quantum harmonic oscillator is
\psi=(\alpha/\pi)^{1/4}e^{-\alpha x^2/2}
where \alpha = \sqrt{ mk/ \hbar^2} .
Compute \Delta x \Delta p known:
Heisenberg Uncertainty Principle:
\Delta p \Delta x >= \hbar/2...
I find 6x10^-15 m for lambda, so the diameter of Am is 1.6x10^-14, so the wave length is less than the diameter, is that mean the wavelength can be exist inside the Am nucleus?