Radioactivity
5 marks · HardTwo isotopes of cobalt are compared in an experiment. Both have atomic number 27, but cobalt-59 is stable, while cobalt-60 is unstable and undergoes beta minus decay.
Explain how the numbers of neutrons affect the stability of cobalt-59 and cobalt-60.
Cobalt-60 has a half-life of 5.3 years. What percentage of a sample will remain after 15.9 years? Show your working.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) Cobalt-59 has 32 neutrons (59-27), which is a stable ratio. Cobalt-60 has 33 neutrons, and this neutron-proton ratio makes it unstable; extra neutrons can make the nucleus too unstable and prone to beta decay. · must show Stability depends on neutron-proton ratio, Cobalt-59 stable with 32 neutrons, cobalt-60 unstable with 33
(b) 12.5% percent · must show Recognise 15.9 / 5.3 = 3 half-lives, Each half-life halves the sample: 100 -> 50 -> 25 -> 12.5%
Common mistakes: Using incorrect exponent in half-life formula; Forgetting cobalt-59 is stable because of neutron number
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Identifies neutrons control nuclear stability by affecting the balance of forces.
A1 (a) · 1 mark · Calculates neutron numbers and links excess neutrons to instability.
M2 (b) · 1 mark · Correctly determines number of half-lives: 15.9/5.3 = 3.
M3 (b) · 1 mark · Halves sample for each half-life (100% > 50% > 25% > 12.5%). (follow-through)
A2 (b) · 1 mark · Correctly states percentage remaining (12.5%). (follow-through)
Worked solution
a) The number of neutrons affects the ratio of neutrons to protons. Cobalt-59 has 32 neutrons (59-27) and is stable. Cobalt-60 has 33 neutrons, which makes the nucleus unstable, so it undergoes beta decay.
b) Number of half-lives = 15.9 / 5.3 = 3. After 1 half-life: 50%. After 2: 25%. After 3: 12.5%. Therefore, 12.5% of cobalt-60 remains.