Consider a particle trapped in a 1D box with zero potential energy with walls at x = 0 and x = L. The general wavefunction solutions for this problem with quantum number, n, are: Vn(x) = sin ( The corresponding energy (level) for each wavefunction solution is: n²h? En 8mL² a) What is the probability of finding the particle between x = L/4 and x = 3L/4 when the particle is in quantum state n = 1, 2 and 3. You can use calculator or a numerical program to do the integral. For people who want to try doing the integral by hand, the following identity will be helpful: sin²(x) = (1 – cos (2x))/2.
Consider a particle trapped in a 1D box with zero potential energy with walls at x = 0 and x = L. The general wavefunction solutions for this problem with quantum number, n, are: Vn(x) = sin ( The corresponding energy (level) for each wavefunction solution is: n²h? En 8mL² a) What is the probability of finding the particle between x = L/4 and x = 3L/4 when the particle is in quantum state n = 1, 2 and 3. You can use calculator or a numerical program to do the integral. For people who want to try doing the integral by hand, the following identity will be helpful: sin²(x) = (1 – cos (2x))/2.
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![Consider a particle trapped in a 1D box with zero potential energy with walls at x = o
and x = L. The general wavefunction solutions for this problem with quantum number,
n, are:
V,6) = sin )
4n(x) =
The corresponding energy (level) for each wavefunction solution is:
n²h?
En
8mL?
a) What is the probability of finding the particle between x = L/4 and x = 3L/4 when
the particle is in quantum state n = 1, 2 and 3.
You can use calculator or a numerical program to do the integral. For people who want
to try doing the integral by hand, the following identity will be helpful: sin²(x) = (1 – cos
(2x))/2.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcb62a11b-b698-47a0-bc01-8f8b792e0947%2F21c7a491-e52d-4459-9013-cff5a7a10c6f%2F9n6pti7_processed.png&w=3840&q=75)
Transcribed Image Text:Consider a particle trapped in a 1D box with zero potential energy with walls at x = o
and x = L. The general wavefunction solutions for this problem with quantum number,
n, are:
V,6) = sin )
4n(x) =
The corresponding energy (level) for each wavefunction solution is:
n²h?
En
8mL?
a) What is the probability of finding the particle between x = L/4 and x = 3L/4 when
the particle is in quantum state n = 1, 2 and 3.
You can use calculator or a numerical program to do the integral. For people who want
to try doing the integral by hand, the following identity will be helpful: sin²(x) = (1 – cos
(2x))/2.
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