Date | May 2021 | Marks available | 3 | Reference code | 21M.2.SL.TZ2.6 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 2 |
Command term | Show that | Question number | 6 | Adapted from | N/A |
Question
All living plants contain an isotope of carbon called carbon-14. When a plant dies, the isotope decays so that the amount of carbon-14 present in the remains of the plant decreases. The time since the death of a plant can be determined by measuring the amount of carbon-14 still present in the remains.
The amount, A, of carbon-14 present in a plant t years after its death can be modelled by A=A0e-kt where t≥0 and A0, k are positive constants.
At the time of death, a plant is defined to have 100 units of carbon-14.
The time taken for half the original amount of carbon-14 to decay is known to be 5730 years.
Show that A0=100.
Show that k=ln 25730.
Find, correct to the nearest 10 years, the time taken after the plant’s death for 25% of the carbon-14 to decay.
Markscheme
100=A0e0 A1
A0=100 AG
[1 mark]
correct substitution of values into exponential equation (M1)
50=100e-5730k OR e-5730k=12
EITHER
-5730k=ln12 A1
ln12=-ln 2 OR -ln12=ln 2 A1
OR
e5730k=2 A1
5730k=ln 2 A1
THEN
k=ln 25730 AG
Note: There are many different ways of showing that k=ln 25730 which involve showing different steps. Award full marks for at least two correct algebraic steps seen.
[3 marks]
if 25% of the carbon-14 has decayed, 75% remains ie, 75 units remain (A1)
75=100e-ln 25730t
EITHER
using an appropriate graph to attempt to solve for t (M1)
OR
manipulating logs to attempt to solve for t (M1)
ln 0.75=-ln 25730t
t=2378.164…
THEN
t=2380 (years) (correct to the nearest 10 years) A1
[3 marks]