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Question 22N.2.SL.TZ0.6b.i

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Date November 2022 Marks available [Maximum mark: 2] Reference code 22N.2.SL.TZ0.6b.i
Level SL Paper 2 Time zone TZ0
Command term Calculate Question number b.i Adapted from N/A
b.i.
[Maximum mark: 2]
22N.2.SL.TZ0.6b.i

Polonium-210 (Po-210) decays by alpha emission into lead-206 (Pb-206).

The following data are available.

  Nuclear mass of Po-210 = 209.93676 u

  Nuclear mass of Pb-206 = 205.92945 u

Mass of the alpha particle = 4.00151 u

(b.i)

Calculate, in MeV, the energy released in this decay.

[2]

Markscheme

(mpolonium − mlead − mα)c2  OR  (209.93676 − 205.92945 − 4.00151)

OR

mass difference = 5.8 × 10−3 ✓

 

conversion to MeV using 931.5 to give 5.4 «MeV» ✓

 

Allow ECF from MP1.

Award [2] for a BCA.

Award [1] for 8.6 x 10−13 J.

Examiners report

Generally, well answered. There were quite a few who fell into the trap of multiplying by an unnecessary c2 as they were not sure of the significance of the unit of u.

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