ALL CHAPTERS COVERED
, 1
Radioactive decay and carbon dating
Exercise 1.1 Radioactive isotopes decay at random, with a fixed probability
of decay per unit time. Over a time interval ∆t, suppose that the probability
of any one isotope decaying is k∆t. If there are N isotopes, how many will
decay on average over a time interval ∆t? Deduce that
N (t + ∆t) − N (t) ≈ −N k∆t,
and hence that dN/dt = −kN is an appropriate model for radioactive decay.
Over a time interval ∆t, N k∆t isotopes will decay. We then have
N (t + ∆t) − N (t) = −N k∆t.
Dividing by ∆t gives
N (t + ∆t) − N (t)
= −N k,
∆t
and letting ∆t → 0 we obtain, using the definition of the derivative,
dN
= −kN.
dt
Exercise 1.2 Plutonium 239, virtually non-existent in nature, is one of the
radioactive materials used in the production of nuclear weapons, and is a
by-product of the generation of power in a nuclear reactor. Its half-life is
approximately 24 000 years. What is the value of k that should be used in
(1.1) for this isotope?
Since N (t) = N (s)e−k(t−s) , half of the isotopes decay after a time T ,
where
N (s + T ) = 12 N (s) = N (s)e−kT ,
1
,2 1 Radioactive decay and carbon dating
i.e. when 12 = e−kT . Thus the half-life T = ln 2/k (as derived in Section
1.1). If T = 24000 then k = ln 2/T ≈ 2.888 × 10−5 .
Exercise 1.3 In 1947 a large collection of papyrus scrolls, including the old-
est known manuscript version of portions of the Old Testament, was found
in a cave near the Dead Sea; they have come to be known as the ‘Dead Sea
Scrolls’. The scroll containing the book of Isaiah was dated in 1994 using
the radiocarbon technique1 ; it was found to contain between 75% and 77%
of the initial level of carbon 14. Between which dates was the scroll written?
We have
N (1994) = pN (s) = N (s)e−k(1994−s) ,
where 0.75 ≤ p ≤ 0.77. Taking logarithms gives
log p = −k(1994 − s),
and so
log p
s = 1994 + .
k
With k = 1.216 × 10−4 this gives (approximately)
−372 ≤ s ≤ −155,
dating the scrolls between 372 BC and 155 BC.
Exercise 1.4 A large round table hangs on the wall of the castle in Winch-
ester. Many would like to believe that this is the Round Table of King Arthur,
who (so legend would have it) was at the height of his powers in about AD
500. If the table dates from this time, what proportion of the original carbon
14 would remain? In 1976 the table was dated using the radiocarbon tech-
nique, and 91.6% of the original quantity of carbon 14 was found2 . From
when does the table date?
If the table dates from 500 AD then we would expect
N (t) = e−k(t−500) N (500),
and so in 2003 we have
N (2003) = e−1503k N (500).
The proportion of 14 C isotopes remaining should there be e−1503k ≈ 83%.
1 A.J. Jull et al., ‘Radiocarbon Dating of the Scrolls and Linen Fragments from the Judean
Desert’, Radiocarbon (1995) 37, 11–19.
2 M. Biddle, King Arthur’s Round Table (Boydell Press, 2001).
, Radioactive decay and carbon dating 3
However, we in fact have 91.6% remaining in 1976. Therefore
N (1976) = 0.915N (s) = N (s)e−k(1993−s) .
Taking logarithms gives
log 0.916
s = 1976 + ≈ 1255;
k
the table probably dates from during the reign of the English King Edward
I, who took the throne in 1270 AD (once the wood was well seasoned) and
had a passion for all things Arthurian.
Exercise 1.5 Radiocarbon dating is an extremely delicate process. Suppose
that the percentage of carbon 14 remaining is known to lie in the range 0.99p
to 1.01p. What is the range of possible dates for the sample?
Suppose that a proportion αp of the original 14 C isotopes remain. Then
αpN (s) = N (t) = e−k(t−s) N (s),
and so
log α + log p = −k(t − s).
It follows that
log p log α
s=t+ + . (S1.1)
k k
Denote by S the value of this expression when α = 1, i.e. S = t + (log p)/k.
For a proportion 0.99p the expression (S1.1) gives
s = S − 82.65,
while for a proportion 1.01p the expression gives
s = S + 81.83
(both correct to two decimal places). Small errors can give a difference of
over 160 years in the estimated date.
, 1
Radioactive decay and carbon dating
Exercise 1.1 Radioactive isotopes decay at random, with a fixed probability
of decay per unit time. Over a time interval ∆t, suppose that the probability
of any one isotope decaying is k∆t. If there are N isotopes, how many will
decay on average over a time interval ∆t? Deduce that
N (t + ∆t) − N (t) ≈ −N k∆t,
and hence that dN/dt = −kN is an appropriate model for radioactive decay.
Over a time interval ∆t, N k∆t isotopes will decay. We then have
N (t + ∆t) − N (t) = −N k∆t.
Dividing by ∆t gives
N (t + ∆t) − N (t)
= −N k,
∆t
and letting ∆t → 0 we obtain, using the definition of the derivative,
dN
= −kN.
dt
Exercise 1.2 Plutonium 239, virtually non-existent in nature, is one of the
radioactive materials used in the production of nuclear weapons, and is a
by-product of the generation of power in a nuclear reactor. Its half-life is
approximately 24 000 years. What is the value of k that should be used in
(1.1) for this isotope?
Since N (t) = N (s)e−k(t−s) , half of the isotopes decay after a time T ,
where
N (s + T ) = 12 N (s) = N (s)e−kT ,
1
,2 1 Radioactive decay and carbon dating
i.e. when 12 = e−kT . Thus the half-life T = ln 2/k (as derived in Section
1.1). If T = 24000 then k = ln 2/T ≈ 2.888 × 10−5 .
Exercise 1.3 In 1947 a large collection of papyrus scrolls, including the old-
est known manuscript version of portions of the Old Testament, was found
in a cave near the Dead Sea; they have come to be known as the ‘Dead Sea
Scrolls’. The scroll containing the book of Isaiah was dated in 1994 using
the radiocarbon technique1 ; it was found to contain between 75% and 77%
of the initial level of carbon 14. Between which dates was the scroll written?
We have
N (1994) = pN (s) = N (s)e−k(1994−s) ,
where 0.75 ≤ p ≤ 0.77. Taking logarithms gives
log p = −k(1994 − s),
and so
log p
s = 1994 + .
k
With k = 1.216 × 10−4 this gives (approximately)
−372 ≤ s ≤ −155,
dating the scrolls between 372 BC and 155 BC.
Exercise 1.4 A large round table hangs on the wall of the castle in Winch-
ester. Many would like to believe that this is the Round Table of King Arthur,
who (so legend would have it) was at the height of his powers in about AD
500. If the table dates from this time, what proportion of the original carbon
14 would remain? In 1976 the table was dated using the radiocarbon tech-
nique, and 91.6% of the original quantity of carbon 14 was found2 . From
when does the table date?
If the table dates from 500 AD then we would expect
N (t) = e−k(t−500) N (500),
and so in 2003 we have
N (2003) = e−1503k N (500).
The proportion of 14 C isotopes remaining should there be e−1503k ≈ 83%.
1 A.J. Jull et al., ‘Radiocarbon Dating of the Scrolls and Linen Fragments from the Judean
Desert’, Radiocarbon (1995) 37, 11–19.
2 M. Biddle, King Arthur’s Round Table (Boydell Press, 2001).
, Radioactive decay and carbon dating 3
However, we in fact have 91.6% remaining in 1976. Therefore
N (1976) = 0.915N (s) = N (s)e−k(1993−s) .
Taking logarithms gives
log 0.916
s = 1976 + ≈ 1255;
k
the table probably dates from during the reign of the English King Edward
I, who took the throne in 1270 AD (once the wood was well seasoned) and
had a passion for all things Arthurian.
Exercise 1.5 Radiocarbon dating is an extremely delicate process. Suppose
that the percentage of carbon 14 remaining is known to lie in the range 0.99p
to 1.01p. What is the range of possible dates for the sample?
Suppose that a proportion αp of the original 14 C isotopes remain. Then
αpN (s) = N (t) = e−k(t−s) N (s),
and so
log α + log p = −k(t − s).
It follows that
log p log α
s=t+ + . (S1.1)
k k
Denote by S the value of this expression when α = 1, i.e. S = t + (log p)/k.
For a proportion 0.99p the expression (S1.1) gives
s = S − 82.65,
while for a proportion 1.01p the expression gives
s = S + 81.83
(both correct to two decimal places). Small errors can give a difference of
over 160 years in the estimated date.
