A radioactive substance has a half life of six months. After how much time will three fourth of the original substance have decayed, leaving one fourth of the initial amount undecayed?

Difficulty: Medium

Correct Answer: Twelve months

Explanation:


Introduction / Context:
This question belongs to nuclear chemistry and radioactive decay. The idea of half life is a standard way of describing how quickly a radioactive substance decays. Many exam questions require you to relate fractions of material remaining to a number of half life periods. In this case, you are asked to find the time when only one fourth of the original sample remains, meaning three fourth of it has decayed.


Given Data / Assumptions:

  • The half life of the radioactive substance is six months.
  • Half life is the time in which half of the substance decays.
  • We want the time required for three fourth of the substance to decay, which is equivalent to one fourth remaining.
  • Assume ideal first order radioactive decay where each half life reduces the remaining amount by half.


Concept / Approach:
Radioactive decay in this context follows an exponential law, but for simple fractions of the original amount, we can think in terms of repeated half life steps. After one half life, half of the substance remains. After two half lives, half of that half remains, meaning one fourth is left. If one fourth is left, then three fourth has decayed. Therefore we simply need to compute the time corresponding to two half life periods and not use the full exponential formula in this simple case.


Step-by-Step Solution:

Step 1: Note that the half life is six months. This means in six months the amount of radioactive substance becomes one half of the initial amount. Step 2: After the first half life of six months, remaining amount equals one half of the original, and one half has decayed. Step 3: After another half life of six months, the remaining amount is half of the previous half, that is one quarter of the original amount. Step 4: At this point, one quarter remains and three quarters have decayed, which matches the condition in the question. Step 5: The total time that has passed for two half lives is six months plus six months equal to twelve months.


Verification / Alternative check:
A more formal approach uses the exponential decay relationship N equal to N0 times (1 divided by 2) raised to the power t divided by T where N is the remaining amount, N0 is the initial amount, t is time and T is the half life. We want N equal to N0 divided by 4, so N divided by N0 equals 1 divided by 4. Set 1 divided by 4 equal to (1 divided by 2) raised to the power t divided by T. Recognise that 1 divided by 4 is (1 divided by 2) raised to the power 2. Therefore t divided by T equals 2, so t equals 2T. Substituting T equals six months gives t equals 12 months. This confirms the answer using the general formula.


Why Other Options Are Wrong:

  • Six months: After one half life, only half the substance has decayed, not three fourth.
  • Ten months: This time does not correspond to an exact integer number of half lives, so the remaining fraction would not be exactly one quarter.
  • Twenty four months: This equals four half lives, after which only one sixteenth of the original amount remains, not one quarter.
  • Eighteen months: This is three half lives, leaving one eighth of the original substance, not one quarter.


Common Pitfalls:
One common mistake is to think that three fourth decay must take three times the half life, but half life steps are multiplicative, not additive in terms of fraction remaining. Another error is to subtract three fourth of the half life rather than thinking in terms of repeated halving. The safest method is to track the fraction remaining after each half life step until you reach the required fraction. This approach works well for questions involving simple fractions such as one half, one quarter and one eighth.


Final Answer:
Three fourth of the substance will have decayed in Twelve months, at which time one fourth of the original amount remains.

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