Charge Is Quantized — Question 8

Section 21-4 · Physics pages 3446–3447

Question 8 Random single-electron arrival produces a noninteger mean but integer outcomes In each of 100 independent trials, a collector gains one electron with probability p = .300 and none otherwise. Let Xi = 1 for capture and 0 for no capture, so deposited charge is Q = −e ∑ i Xi. Use Var(Xi) = p(1 −p) and add variances for independent trials. trial −e 0 .300 .700 Schematic; labels, not drawing lengths, specify the data. (a) Find the mean deposited charge, its variance and standard deviation in units of e. (b) Find the probability of no deposited charge. Explain how an average charge of −.300e per trial is consistent with integer electron transfer.
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Question 8 - Solution Random single-electron arrival produces a noninteger mean but integer outcomes trial −e 0 .300 .700 (a) Method and calculation ⟨Q⟩= −100pe = −30e, Var(Q) = 100e2p(1 −p) = 21e2. σQ = √ 21e ≈4.58258e. (b) Interpretation and further calculation P(Q = 0) = (1 −p)100 = .7100 ≈3.23448 × 10−16. Q/e ∈{0, −1, . . . , −100}, ⟨Qi⟩/e = −.300. An ensemble mean is a probability-weighted average, not the charge transferred in a particular trial. Physical check The square-root variance rule depends on independence; correlated captures would change the fluctuation size.

Question 8

Random single-electron arrival produces a noninteger mean but integer outcomes

In each of 100 independent trials, a collector gains one electron with probability p=.300p=.300 and none otherwise. Let Xi=1X_i=1 for capture and 0 for no capture, so deposited charge is Q=−e∑iXiQ=-e\sum_iX_i. Use Var⁡(Xi)=p(1−p)\operatorname{Var}(X_i)=p(1-p) and add variances for independent trials.

Diagram shown in the original SVG worksheet.

  1. Find the mean deposited charge, its variance and standard deviation in units of ee.

  2. Find the probability of no deposited charge. Explain how an average charge of −.300e-.300e per trial is consistent with integer electron transfer.

Question 8 - Solution

Random single-electron arrival produces a noninteger mean but integer outcomes

Diagram shown in the original SVG worksheet.

(a) Method and calculation

⟨Q⟩=−100pe=−30e,Var⁡(Q)=100e2p(1−p)=21e2.\langle Q\rangle=-100pe=-30e,\quad\operatorname{Var}(Q)=100e^2p(1-p)=21e^2.σQ=21e≈4.58258e.\sigma_Q=\sqrt{21}e\approx4.58258e.

(b) Interpretation and further calculation

P(Q=0)=(1−p)100=.7100≈3.23448×10−16.P(Q=0)=(1-p)^{100}=.7^{100}\approx3.23448\times10^{-16}.Q/e∈{0,−1,…,−100},⟨Qi⟩/e=−.300.Q/e\in\{0,-1,\ldots,-100\},\quad\langle Q_i\rangle/e=-.300.An ensemble mean is a probability-weighted average, not the charge transferred in a particular trial.

Physical check

The square-root variance rule depends on independence; correlated captures would change the fluctuation size.