Tuesday, 26 November 2013

Interesting SPM 2013 Question on Half-Life

1) SPM 2013 Physics Paper 1 has an interesting question on half-life calculation.

Q 48 reads:

"25% of Gallium-65 atoms have decayed after 15 minutes. What is the half-life of Gallium-65?

A 30.0 minutes

B 23.4 minutes

C 15.0 minutes

D 7.5 minutes

(Shouldn't there be another choice, say:

E 36+ minutes?)

My Comments (based on info given in the Q48):

Half-life of Gallium-65 based on info given in the Q48:
25% have decayed means 75% have not decayed. And that means, its half-life > 15 minutes. So was the half-life A, B or none of the 4 given choices? Do you not agree that the half life should be calculated as follows:

The 1/4-life Method:

At the end of 15 minutes (1st 1/4-life):   
a) amount decayed = 25% = ¼ of original amount
b) amount undecayed = 75% = ¾ of original amount

At the end of 30 minutes (2nd 1/4-life):  
a) amount decayed   ¼ x ¾ of original amount = 3/16 of original amount
b) total amount decayed = (¼ + 3/16) of original amount = 7/16 of the original amount
c) total amount undecayed = (1 – 7/16) of original amount = 9/16 of the original amount

To decay to ½ (or 8/16) of original amount:
a)      A further 1/16 is to be decayed
b)      1/16 = [(1/4) x (9/16)] x (4/9)
c)      Time needed = [15 minutes] x (4/9) =  6.67 minutes

Therefore, half-life = (15 + 15 + 6.67) minutes
                               = 36.67minutes

Answer (30 minutes) cannot be the acceptable answer because in 30 minutes only 7/16 and not 1/2 (or 8/16) of the radioactive substance has decayed. Conceptually, A is a wrong answer. Q48, to me, has no acceptable answers. This is also confirmed by logarithm method of calculating half-life as follows:

Logarithm Method to Calculate Number of Half-Lives:

Let:      n = number of half-lives
            Ao = Original amount of radioactive substance (or original level of radioactivity)
            Ac = Current amount of radioactive substance (or current level of radioactivity)

Then,    (1/2)n x Ao = Ac           
                        (1/2)n = Ac/Ao
                        (1/2)n = 0.75Ao/Ao = 0.75

log both sides:      log (1/2)n = log (0.75)
 Hence,       n  = log (0.75) ÷ log (1/2) = 0.415037499 half-life

To Calculate the Half-Life, T1/2

Let       T1/2 = Half-life time
            To = Time as at original amount of radioactive substance
(or original level of radioactivity)
            Tc = Time as at current amount of radioactive substance
(or current level of radioactivity)
            n = number of half-lives

Then,    T1/2 = (Tc - To) ÷ n = 15 minutes ÷ n = 15 min ÷ [log (0.75) ÷ log (1/2)]

            T1/2 = 36.14131259

            T1/2 = 36.1 minutes

Based on facts given in Q48, logarithm method shows a half-life 36.1 minutes.

But online searches (please click here or here) show that isotopes Gallium-65  actually have a half-life of 15.2 minutes

So, which is the examiners' preferred answer and why?

Isn't this question both interesting and bewildering?! :)

In conclusion: Overall, except for Q48, Paper 1 is still a good paper - of respectable standard! :)

(10.4.2014: The examiners should have replaced the words "Gallium-65" in Q 48 with the words "isotopes-X". Otherwise, it's like a biology question on man that talks about the man having ovaries, uterus, etc. that makes the candidates wonder whether it is a question about man or something else. The more well-read the students, the more disadvantaged he/she would be - that is ridiculous!)


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