AQA A-Level Physics Paper 3 (3BB), June 2025: Question 2

Question Answers Additional comments/Guidance Mark AO

02.1 Max one from Expect to see 74 (× 109) − 58 (× 109) (= 16 × 109) 2 AO2

• Reads off activities of 74(× 109) and 58 (× 109) from Condone consistent power of ten errors where attempt

the graph (and calculates the difference) (1) to find difference is seen

Condone one other error in read offs where their

difference is calculated.

Expect to see one of

• Use of A = −λN to find number of nuclei at t=0 / 74×109

• (N=) (= 2.64 × 1016) (a)

t =24 / difference between t=0 and t =24 (2) 2.8×10−6

58 × 109

• (N=) (= 2.07 × 1016) (b)

2.8 ×10−6

16 × 109

• (N=) −6 (c)

2.8 ×10

• 74(× 109) − 58 (× 109) = 2.8 (× 10-6) × N (d)

• 16 (× 109) = 2.8 (× 10-6) × N (e)

• Use of N = N 𝑒𝑒−𝜆𝜆𝑡𝑡 (3)

0 16 16

• Finding difference of 2.64 × 10 and 2.07 × 10

seen (f)

−𝜆𝜆𝑡𝑡 difference in activity

• Use of (∆)N = N0 (1 − 𝑒𝑒 ) (4) • Condone one error in -6 (g)

2.8 ×10

condone power of ten errors in (d) and (e) and (f)

For Use of in (3) is by substitution of N (or N0) and t = 24

• determines area under 99Mo graph between t = 0 and (× 602) , condone N (1.82 x1016) from A = 51 GBq.

t = 24 hours. (5) For (3) N = 2.64 (× 1016) / N= 2.07 (× 1016) t = 24 (× 602)

For (4) N = 2.64 (× 1016) t = 24 (× 602)

allow area of a trapezium: expect base 24 × 3600 and

parallel sides 74 × 109 and 58 × 109

5.7 × 1015

allow counting squares: expect 325 to 400 small squares

(13 to 16 large squares) x by area of one square

for MP2 accept > 2 sf, rounding to 5.7 × 1015

Must see MP1 to award MP2

02.2 Method 1: for number in range 3.0 to 3.7 × 1015 3 1 × AO2

calculates number of 99mTc (that have decayed between t = 2 × AO3

0 and t = 24) 1 For allow use of (0.9 ×) 6 × 1015 − their number

99m of 99mTc

use of (0.9 ×) their 02.1 − their number of Tc (that have

decayed between t = 0 and t = 24) 2 for 3 must see use of 0.9 in a valid difference

calculation

number of 99mTc extracted in range 1.4 to 2.4 × 1015

for 1 evidence for half-life of 6(.0) hours seen in

working; expect λ = 3.2 × 10−5 (s−1 ) based on half-

Method 2: use of recognisable half-life to determine the life = 2.2 × 104 s

decay constant for 99mTc ln 2

1 allow substitution of half-life in hours, eg λ =

6(.0)

use of activity at t = 24 hours and a decay constant to for substitution of A in range 50(× 109) to

99m 2

determine number of Tc extracted 2 ( 9 )

9 51×10

52(.2) (× 10 ), eg N =

their λ

allow PoT on activity in MP2

number of 99mTc extracted in range 1.50 to 1.63 × 1015 15

3 for allow 2 sf in range 1.5 to 1.6 × 10

Where no other mark is scored, Max 1 for:

attempt to find area (under 99mTc curve)

• 215 to 255 small squares (or 8 to 10 large)

• allow area of triangle and trapezium

• finds area of one small square = 1.4 × 1013

Question Answers Additional comments/Guidance Mark 11 AO

02.3 Mark Criteria In each area, a partial response covers one bullet point 6 2 × AO2

while a detailed response covers at least two.

All three areas covered in some detail. Area 1: How images obtained (MUGA scan) 4 × AO3

66 marks can be awarded even if there is an error • (99m)Tc is used as a tracer OR (99m)Tc is a gamma-

and/or parts of one aspect missing. emitter (do not award if annihilation mentioned)

All three areas covered, at least two in detail. • the imaging device is a gamma camera

5 Whilst there will be gaps, there should only be an (mentioning a detector / scanner insufficient)

occasional error. • details of gamma camera (collimator,

photomultiplier, dynodes)

Two areas successfully discussed, or one discussed treat as neutral: (99m)Tc quickly removed from body (in

and two others covered partially. Whilst there will be about 24 hours) / has a short (effective) half-life

4 several gaps, there should only be an occasional

Area 2: Common advantages

error. • both are non-invasive

One area discussed and one discussed partially, or all • both generate (2D) images that provide an

3 three covered partially. There are likely to be several animated view of the functioning heart or wtte

errors and omissions in the discussion. • no need to anaesthetise patient

Only one area discussed or makes a partial attempt at • both are outpatient procedures

2 two areas. • patients do not have to remain still / allow ‘both

may be used during stress testing’

1 Only one area covered and that partially. 'can be used if patient has pacemaker' is neutral

Area 3: Strength and weakness

0 No relevant comments • MUGA provides (both) quantitative (and

qualitative) information / MUGA is more accurate /

TTE provides only qualitative information / MUGA

image intensity is related to number of red blood

cells in region

• Interpretation of TTE images may be subjective /

not reproducible

• TTE lower cost / allow ‘takes less time’

allow reverse argument

• TTE uses non-ionising radiation / no side effects /

MUGA uses ionising radiation.

’MUGA produces higher resolution images’ is neutral

Total 11