In Example 4.2. neither|Ψ|2nor|Ψ|are given units—only proportionalities are used. Here we verify that the results are unaffected. The actual values given in the example are particle detection rates, in particles/second, ors-1. For this quantity, let us use the symbol R. It is true that the particle detection rate and the probability density will be proportional, so we may write|Ψ|2= bR, where b is the proportionality constant. (b) What must be the units of b? (b) What is|ΨT|at the center detector (interference maximum) in terms of the example’s given detection rate and b? (c) What would be|Ψ1|,|Ψ1|2, and the detection rate R at the center detector with one of the slits blocked?

Short Answer

Expert verified

a)Units of b iss/m

(b) The detection rate at the center detector and b is10bs1/2

(c) The detection rate R is25s-1

Step by step solution

01

Dimensional analysis.

(a)

The dimensional analysis comes to the rescue, as we are told that the units of measurement areψ2 is and1m that of R is 1s. Since the relation between ψ2and R isψ2=bR

Therefore, the units of b ares/m

02

Detection rate.

(b)

In example 4.2, we were given that the detection rate (R) is 100, hence, we can directly findψT it from the relations given in part (a).

|ψ1|2=(100b)s1

|ψ1|=(10b)s1/2

03

 Step 3: Detection rate R.

(c)

The wave amplitude will be halved with just a slit open, thus we can find the new detection rate R using the relation derived in section b).

|ψ1|=12|ψT|=12×(10bs1/2)

|ψ1|2=(25b)s1|ψ1|2=R1b

Comparing it we get,

R1=25 s-1

Result.

(a)Units of b iss/m

(b) The detection rate at the center detector and b is10bs1/2

(c) The detection rate R is25s-1

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