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A one-dimensional infinite well of length 175 pm contains an electron in its third excited state. We position an electron-detector probe of width 3.50 pm so that it is centered on a point of maximum probability density. (a) What is the probability of detection by the probe? (b) If we insert the probe as described 500 times, how many times should we expect the electron to materialize on the end of the probe (and thus be detected)?

User Herve
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1 Answer

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Answer:

a. 0.04, b = 40

Step-by-step explanation:

a. The probability of detection at x is

P(x) = ψ²ₙ(x) dx

P(x) = [(2/L)^1/2 sin((πn/L) x)]² dx

P(x) = 2/Lsin²(πn/L x) dx

Given that

n =31, L = 175pm and dx = 3.5 pm

X = L/2 = 175/2 = 87.5 pm

P(x = L/2) = 2/L sin²(3π/L . L/2) dx

P(x = L/2) = 2/L sin²(3π/2) dx

P(x = L/2) = 2/L dx = (2/175) * 3.5 = 0.04

The probability of detection is 0.04

b. N = 500 independent insertions

N = Np

N = 500 x 0.04 = 40

The electrons should materialize 40 times

User Bruno Grieder
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