- #1
Anonymous1783
- 3
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Hi, (first ever post)
I have the following three problems
1) A particle is in the ground state of an infinite well from 0<x<a. Find the probability that it is in the middle half of the well.
I think I have managed this, its just the integral of the wave function squared then I put in the values 3a/4 and a/4 to get the probability. It all cancels out nicely to get 0.82 which seems about right
psi = (2/a)^0.5 * sin(pi*x/a)
psi^2 = (2/a) * sin^2(pi*x/a)
Integral = ((2/a)((x/2)-((a/4pi)sin(2pix/a)
then put in the limits 3a/4 and a/4 and get 0.82
Is this the right method
2) A particle is in second state n=3 of an infinite well from -a/2<x<a/2. Find the position of the two nodes.
psi = (2/a)^0.5 * sin(pi*x*3/a)
probability density = psi^2 = (2/a) * sin^2(pi*x*3/a)
At the nodes i thought this must be equal to zero
so (2/a) * sin^2(pi*x*3/a)=0
therefore the nodes must be at x=a/3 and x=-a/3
3) This is what I am stuck on. Consider a well of finite depth that is just deep enough to contain three states. Sketch n=3 wave function and give the two values of the two nodes in this case.
I have sketched it from a textbook but I don't know how to find the nodes because I am confused about what the wave function is. Would it just be Asin(kx)+Bcos(kx). Any thoughts appreciated.
Thanks
I have the following three problems
1) A particle is in the ground state of an infinite well from 0<x<a. Find the probability that it is in the middle half of the well.
I think I have managed this, its just the integral of the wave function squared then I put in the values 3a/4 and a/4 to get the probability. It all cancels out nicely to get 0.82 which seems about right
psi = (2/a)^0.5 * sin(pi*x/a)
psi^2 = (2/a) * sin^2(pi*x/a)
Integral = ((2/a)((x/2)-((a/4pi)sin(2pix/a)
then put in the limits 3a/4 and a/4 and get 0.82
Is this the right method
2) A particle is in second state n=3 of an infinite well from -a/2<x<a/2. Find the position of the two nodes.
psi = (2/a)^0.5 * sin(pi*x*3/a)
probability density = psi^2 = (2/a) * sin^2(pi*x*3/a)
At the nodes i thought this must be equal to zero
so (2/a) * sin^2(pi*x*3/a)=0
therefore the nodes must be at x=a/3 and x=-a/3
3) This is what I am stuck on. Consider a well of finite depth that is just deep enough to contain three states. Sketch n=3 wave function and give the two values of the two nodes in this case.
I have sketched it from a textbook but I don't know how to find the nodes because I am confused about what the wave function is. Would it just be Asin(kx)+Bcos(kx). Any thoughts appreciated.
Thanks