Probability of finding a particle in a 1D box

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SUMMARY

The discussion focuses on calculating the probability of finding a particle in a one-dimensional box of length 1 nm using the wave function ψ = (2/L)½ sin(πx/L). The probability density function is derived as ψ² = (2/L) sin²(πx/L), and the probability between x = 0.55 nm and x = 0.65 nm is computed through integration, yielding a result of approximately 0.1796. Participants confirm that using nanometers consistently in calculations is essential for accuracy, and suggest verifying results by calculating probabilities over larger intervals for cross-checking.

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Lily Wright
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Homework Statement


If a one-dimensional box is 1 nm long, what is the probability of finding the particle between the following limits?
(a) x = 0 nm and x = 0.05 nm
(b) x = 0.55 nm and x = 0.65 nm

Homework Equations


ψ = (2/L)½ sin(πx/L)

The Attempt at a Solution


(I do chemistry and I'm really terrible at physics so apologies if this makes no sense) So that's the equation I was given. But then the probability is ψ2 which = 2/L sin2 (πx/L).
And to find the probability between x = 0.55 nm and x = 0.65 nm I need to integrate this between those values. So that's what I've done putting L = 1 (but it's in nm?)
0.650.55 ψ2(x)dx and then I get (πx−(sin(2πx)/2))π + C and then the rest is pretty difficult to type out but I basically end up with an answer of 0.1796. Any help would be much appreciated.
 
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Lily Wright said:
So that's what I've done putting L = 1 (but it's in nm?)
You can plug in "1nm", but working in nanometers everywhere will give the correct result as well.

The answer looks reasonable. As a cross-check, you can calculate the probability between 0 nm and 0.5 nm or between 0 nm and 1 nm as those probabilities are easy to get in a direct way.
 

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