Wave function for Hydrogen and Probability

AI Thread Summary
To calculate the probability of finding an electron in the 2 s state of a hydrogen atom at a distance less than 3.00 a from the nucleus, integration of the probability density function is necessary. The probability density is derived from the square of the wave function, represented as P(r) = Ψ(r)^2. The integration should be performed from 0 to 3.00 a to determine the probability. Additionally, the result must be multiplied by the volume of a sphere with a radius of 3.00 a to obtain the actual probability. This approach will yield the desired probability value.
Twigs
Messages
24
Reaction score
0
Hey, if anyone can throw in some thoughts I am a little lost. Not sure If I need to integrate, or what. Thanks for any help.

The wave function for a hydrogen atom in the 2 s state is:(attachment)


I need to Calculate the probability that an electron in the 2 s state will be found at a distance less than 3.00 a from the nucleus.
 

Attachments

  • render.gif
    render.gif
    1.2 KB · Views: 496
Physics news on Phys.org
Of course you need to inegrate.The Probability is the integral of the probability density...

\mathcal{P}=\iiint_{\mathcal{D}} \left| \Psi \left(\vec{r},t\right) \right|^{2} \ dV

is the probability of finding the electron (infinite mass nucleus) in the domain \mathcal{D}\subseteq \mathbb{R}^{3}

Daniel.
 


To calculate the probability, you will need to use the probability density function, which is given by the square of the wave function. So, for the 2 s state of a hydrogen atom, the probability density function would be:

P(r) = Ψ(r)^2

Where r is the distance from the nucleus and Ψ(r) is the wave function. Since you are looking for the probability at a specific distance (less than 3.00 a), you will need to integrate the probability density function from 0 to 3.00 a.

∫P(r)dr = ∫Ψ(r)^2dr

You can use any integration method to solve this integral. Once you have the value, it will give you the probability of finding the electron at a distance less than 3.00 a from the nucleus. Remember, the probability density function gives the probability per unit volume, so you will need to multiply the result by the volume of a sphere with radius 3.00 a to get the actual probability.

Hope this helps!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top