Probability that the electron is found at a distance greater than r

hansbahia
Messages
54
Reaction score
0
The problem given is to calculate the probability that the electron is found at a distance greater than r=2a0 from the center of the hydrogen atom in its ground state.

I don't understand what the problem is asking. I don't understand what form of the wavefunction I should use.

I know that to calculate the probability is P(r)= abs value (ψ(r))^2 or P(r)= ψ*(r).ψ(r)

but what wave function should i use in this problem? should i use e^(r/a0)/sqrt(pi.a0)?

i know that at the ground state n=1 in which makes n,l=0
 
Physics news on Phys.org
Just take the ground state wavefunction and integrate the square from 2a0 to infinity, P(r>2a_0) =\int_{2a_0}^\infty \psi^*(r) \psi (r) dr
 
Thank you!
 
I have a question, in order to solve the same problem but if we use the wave function of the Radial functions of the Hydrogen atom is $$R10= 2/(sqrt(a0^3))*e^(-r/a0)$$ ?
 
Last edited:
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Back
Top