3n s(2x – *), find 4normalized, the normalized wave function for a 1-dimensional particle- in-a-box where the box boundaries are at x=0 and x=2n. The potential energy is zero when 0
3n s(2x – *), find 4normalized, the normalized wave function for a 1-dimensional particle- in-a-box where the box boundaries are at x=0 and x=2n. The potential energy is zero when 0
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Question
![3n
(2x –), find Ynormalized, the normalized wave function for a 1-dimensional particle-
Given 4 = cos
in-a-box where the box boundaries are at x=0 and x=2. The potential energy is zero when
0<x< 2n and ∞ outside of these boundaries.
Y is already normalized. That is, Y = 4,
normalized.
4'normalized = 1 2 cos (2x – )
пот
2
sin (2x -)
-
4normalized
2
4normalized
sin ( 2x
2
-
none are correct](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20258784-e3f1-4063-8300-9502a1f690d1%2F990329f6-5410-407a-9c18-6430dc1e3c49%2F6hhr9so_processed.png&w=3840&q=75)
Transcribed Image Text:3n
(2x –), find Ynormalized, the normalized wave function for a 1-dimensional particle-
Given 4 = cos
in-a-box where the box boundaries are at x=0 and x=2. The potential energy is zero when
0<x< 2n and ∞ outside of these boundaries.
Y is already normalized. That is, Y = 4,
normalized.
4'normalized = 1 2 cos (2x – )
пот
2
sin (2x -)
-
4normalized
2
4normalized
sin ( 2x
2
-
none are correct
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