36

Physics Factsheet www.curriculumpress.co.uk

April 2002

Number 36

Comparison of capacitor discharge and radioactive decay

num ber of undecayed nuclei

number of undecayed nuclei

It is very common that in nature quantities die away with time or decay. It makes total sense that the more there is of something the faster it will decay. In radioactivity the more undecayed nuclei there are the greater the probability that one will decay we write this as A=λ N i.e. the activity is proportional to the number of nuclei that are present. Of course as more of the nuclei decay the rate at which they will now decay falls off. This means that the activity or rate of decay takes the following shape:

One of the characteristics of an exponential decay is that it has a constant half-life. This means that the quantity that is decaying falls by a constant ratio in equal time intervals. The half-life time is the time that it takes for the quantity to decay to half the previously chosen value – it really doesn’t matter where you start from - the half-life will always be the same. Worked Example: the activity of a sample of a radioactive material is 500 Bq at a certain time and has fallen to 62.5 Bq after 20 minutes. Calculate the half-life of the radioactive material. Answer:

as the number of undecayed nuclei falls the rate of decay falls too so curve tapers off

in one half-life the activity
250 Bq in two half-lives the activity
125 Bq

as the num ber of undecayed nuclei falls the rate of decay falls too so the curve tapers off

in three half-lives the activity
62.5 Bq three half-lives = 20 minutes ⇒ half-life = 20/3 minutes = 1200/3 seconds = 400 seconds Exam Hint: don’t forget that in radioactivity experiments the count rates need to be corrected for background time

General exponential decay equation

general decay equation X = X0e−kt

radioactivity k = λ (decay constant)

count rate X=R

activity X=A

capacitor discharge k = 1/RC (time constant)

number of nuclei X=N

current X=I

1

charge X=Q

voltage X=V

Physics Factsheet

Comparison of capacitor discharge and radioactive decay.

www.curriculumpress.co.uk

A capacitor stores electric charge. If its plates are connected by a resistor a current flows and the capacitor discharges.

The atoms in a radioactive substance have unstable nucleii. They do not, however, all decay together - the decay is a random process with nuclei of atoms seperately decaying in a random order.

C

R Initial Charge Q0 The graph of number of active nuclei (N) against time is shown below

The graph of charge (Q) against time is shown below. charge (Q)

no. of active nuclei remaining

½ No

¼ Qo

½ Qo

One half life T ½

¼ No

N0

Q0

One half life T ½

time Two half lives

time Two half lives

−t/RC

The equation of the graph is N = N0e

The equation of the graph is Q = A0e

The graph of voltage against Q time is of the same form, because V =C . The equation of the voltage

−λt

−λt

or A = A0 e

−t/RC

graph is V = V0e

For both graphs λ is known as the decay constant and it can be shown that λ is related to the half life by the equation: λ = Time 0 T½ 2T ½ 3T ½ nT½

Charge on capacitor plates Q0 ½ Q0 ¼ Q0 1 8 Q 0

Time 0 T½ 2T ½ 3T ½

(½)n Q0

nT½

ln2 = 0.693 ≈ 0.7 T½ T½ T½

Number of atoms still active N0 ½ N0 ¼ N0 1 8 N 0 (½)n N0

For different radioactive sources the half lives can vary enormously, from a fraction of a second to thousands of years. α , β and γ decays all follow the same equation.

The quantity RC is sometimes referred to as the time constant (τ)

For both cases the size of the rate of decay against time shows a very similar shaped graph to the ones above. The decay rates are given by the following equations. -λτ = − Q Decay rate = − 1 Q0e RC RC For the capacitor case the rate of decay is known as the discharge current.

Decay rate = −λ N0e

-λτ

= −λΝ

For radioactivity the rate of decay is known as the activity of the source.

Discharge current

Activity of source

λ No

Q0/RC

time

time

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Comparison of capacitor discharge and radioactive decay.

Physics Factsheet www.curriculumpress.co.uk

Worked Examples:

24

A 5 µF capacitor is charged fully and the potential difference between its plates is 12V. It is then discharged throuygh a 2MΩ resistor. Find each of the following: (a) (b) (c) (d)

the time contant; the decay contant; the half life for the discharge; the initial discharge current.

(a)

Time constant

(b)

Decay constant =

1 RC

=

1 10

=

0.1 s

=

1n2 0.1

=

7s

(c)

Half life

(d)

= = =

The radioisotope 11Na has a half life of 15 hours. For a freshly prepared sample of mass 1 gram, find each of the following: (a) the number of active atoms initially. (b) the number of active atoms after (i) 15h and (ii) 60h. (c) the initial activity of the source. 23 (Avagadro's number = 6.02 × 10 atoms per mole.) (a)

24 1g is 1 mol of 11Na 24 So 1 × 6.02× 1023 = 2.5 × 1022 (= N0) 24

(b)

(i)

After one half-life, N = 1.25 × 1022

(ii)

60h is 4 half-lives

RC 2 × 106 × 5 × 10-6 10 s

22

2.5 × 10 = 1.56 × 1021 4 2 ln2 A0 = λ N0 = × N0 T 1/2 22 = 0.69 × 2.5 × 10 15 × 60 × 60 N =

(c)

Initial discharge current = Q 0 = CV0 RC RC = V0 (cancelling C) R 12V = 2MΩ

= 3.2 × 1017 Bq Exam Hint: Initial values of quantities are usually denoted by the subscript "0" (representing at time = 0); hence Q0 , V0 , N0 etc.

= 6 × 10-6 A or 6 µ A

Exam workshop This is a typical student's answer to an exam question. The comments explain what is wrong with the answers and how they can be improved. The examiner's answer and comments are given below.

(c) the initial activity, in Bq, of the sample of radium.

∆N ∆t =-λN


A sample of radium contains 3.5 × 1021 radioactive nuclei. Radium has a half-life of 1.6 × 103 years. Calculate: (a) the time taken for the number of nuclei to decay to 1.6 × 1018; [5] N = N0 e−λt
1.6 × 1018 = 3.5 × 1021 e-1600t 

= 1.6 × 103 × 3.5 × 1021  = 5.6 × 1024 Bq 

-7.69 = 4.8 × 10-3 years
ecf -1600

3/5 Examiner’s comments and answers: The student’s layout for each of these answers is exemplary and little needs to be changed in order to make the answers into very good ones.

The student has only made one mistake here; s/he has not converted the half-life into the decay constant λ ( = 0.69/T1/2). Having made this mistake, the remainder of the answer is correct and is not further penalised.

In part (a) calculating the decay constant using λ = 0.69/T1/2 gives a value of 4.3 × 104 year-1. Substituting this into the equation N = N0e−λt gives 1.6 × 1018 = 3.5 × 1021 e−43000t Dividing by 3.5 × 1021 and taking logs to get rid of the “e” gives: ln(1.65 × 1018/ (3.5 × 1021)) = -4.3 × 10-4 × t t = 1.8 × 104 year

(b) the number if nuclei that will be present 1000 years from the present; [4] N = N0 e−λt
= N0 e−1600000  = 3.5× 1021 × e−1600000  = 3.5× 1021 × 0 = 0 

1/3

Two mistakes here - the usual failure to change the half-life into the decay constant and also the student's unwillingness to work in seconds. Even with error carried forwards, no credit could be given for the answer in Bq - since Bq are identical to counts per second. Had the question not specifically sought an answer in Bq, the student might well have gained an extra mark for using year-1 in this final answer.

1.6 × 1018 -1600t
(with error carried forward [ecf]) 3.5 × 1021 = e t=

[3]

In part (b) use of the value for ? and the equation N = N0e−λt (again) gives a value for N of 2.3 × 1021 nuclei.

1/4

The student has made the same mistake in this part and is penalised for it again - which may seem a bit harsh, but it is another part of the question. S/he has then gone on to equatie the power of e to zero - which it is not! It looks like the student's calculator is set to a rounding mode which does not give answers as powers of ten - since the student's answer is very small, it has rounded it to zero. Be very careful with the mode to which you set your calculator ("SC" for scientific is probably the best).

In part (c) the decay constant must be given in s-1 to give an activity in Bq. λ = 0.69/(1.6 × 103 × 365 × 3600) = 3.3 × 10-10 s-1 ∆N = -λ N = -3.3 × 10-10 × 3.5 × 1021 = -1.1 × 1012 Bq. ∆t (the minus sign simply tells us that this is a decay – the gradient of the graph of N against t is negative).

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Comparison of capacitor discharge and radioactive decay.

Physics Factsheet www.curriculumpress.co.uk

Questions

Answers 1 (a) R = R0 e−λt
0.75 = 3.75 e−λt ln (0.75/3.75) = −λ t
= -0.69t/5500 t = 12800 years
a)

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1. (a) Living matter contains the 6C isotope which begins to decay when the matter dies. The half life of the isotope is estimated to be 5500 years. A sample of wood from an ancient building has an activity. which registers 0.75 counts per minute, on a ratemeter. If a sample of newly cut wood of the same mass gives a reading of 3.75 counts per minute, how old is the specimen? (b) If a capacitor of capacitance 1F were available, find the resistance required if the discharge half life is to be the same as that of the isotope in part (a). 2.

(b)

T1/2 = 5500 (years) × 3600 (seconds) × 24 (hours) × 365 (days) = 1.73 × 1011 s


The longest half life for a capacitor discharge in common use is that used in a digital clock. If a certain model of clock is disconnected from its supply it will still show the correct time if reconnection is made within 20 minutes, because the electronic system can still operate only half voltage. If a capacitor and a 750k Ω resistor are connected in parallel with the supply find the capacitance of the capacitor. Ignore the resistance of the clock.)

C

1 0.69 =
RC T 1/2

R = (1.73 × 1011)/ (0.69 × 1 (F)) = 2.5 × 1011 Ω


2

V = V0 e−t/RC
½ = e−t/RC -ln(½ ) = -t/RC ln2 = t/RC
since t = 1200 s
C = 1200/ (0.69 × 750 × 103) = 2.3 × 10-3 F
(= 2.3 mF)

3.

750k Ω

Similarity

Capacitor voltage Radioactive no. of atoms (or current, charge) (or activity, count rate)

Exponential decay V = V0e−t/RC
with time, t

CLOCK

Initial value
at t = 0 V0 3.

Describe the similarities between the mathematical explanations of capacitor discharge and radioactive decay.

Time constant



N = N0e−λt N0

1/RC (R = discharge λ, where circuit resistance, λ = decay constant C = capacitance)

Acknowledgements: This Physics Factsheet was researched and written by Alfred Ledsham The Curriculum Press,Unit 305B, The Big Peg,120 Vyse Street, Birmingham, B18 6NF Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136

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