DP Chemistry (last assessment 2024)

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Question 20N.3.hl.TZ0.17d

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Date November 2020 Marks available [Maximum mark: 2] Reference code 20N.3.hl.TZ0.17d
Level hl Paper 3 Time zone TZ0
Command term Calculate Question number d Adapted from N/A
d.
[Maximum mark: 2]
20N.3.hl.TZ0.17d

Technetium-99m is the most commonly used isotope for diagnostic medicine.

(d)

Technetium-99m has a half-life of 6.03 hours. Calculate the amount of 1.00×10-11 mol of technetium-99m remaining after 48.0 hours.

[2]

Markscheme

Alternative 1:

«N=N0(0.5)tt1/2=»1.00×10−11×0.548.06.03 ✔

«N=»4.02×10−14 «mol» ✔


Alternative 2:
«λ=ln 26.03=»0.115«hr−1» ✔

«N=N0e−λt=1.00×10−11×e−0.115×48=»4.01×10−14 «mol» ✔


Award [2] for correct final answer.

Examiners report

The question on half-life was very well answered and nearly all scored full marks, often via different methods of calculation of the amount of technetium-99m remaining after a period of 48 HRS.

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