Finding the probability of an electron in the forbidden region

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Homework Help Overview

The discussion revolves around calculating the probability of finding an electron in a classically forbidden region for a hydrogen atom in its first excited state. The problem involves quantum mechanics concepts and the associated mathematical framework.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to integrate over three-dimensional space rather than just one variable. There are questions about the proper expression for the volume element in spherical coordinates and the significance of the variable used for charge.

Discussion Status

Some participants have provided guidance on the integration process and have pointed out potential issues with the original poster's approach. There is an ongoing exploration of the mathematical details and assumptions involved in the problem.

Contextual Notes

Participants note the importance of correctly interpreting variables and limits in the context of quantum mechanics. There is also a mention of the original poster's notation choices and their implications for the calculations.

reminiscent
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Homework Statement


For the hydrogen atom in the first excited state, find the probability of finding the electron in a classically forbidden region.

Homework Equations


E1 = -Z2μe'4/2n2ħ2
ψ200 =
∫ψ(r)2dr from the limit to ∞

The Attempt at a Solution


https://imgur.com/a/fWIMq

I feel like this answer is too weird... I would like my work to be checked on.
 
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Bump.
 
Do I see an ##e^{14}## in your ##E## ?
 
reminiscent said:
I feel like this answer is too weird... I would like my work to be checked on.
Your work looks pretty good. However, you need to integrate over 3 dimensional space, not just over ##r##. How would you express the volume element ##dV## in terms of spherical coordinates?

What is ##Z## for hydrogen?

For the ##r## part of the integration, let the variable of integration be ##\sigma = r/a_0##. The lower limit will then become a simple number.
 
BvU said:
Do I see an ##e^{14}## in your ##E## ?
I think the OP is using ##e'## for the proton charge in order to distinguish it from the base of the natural log function.
 
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