Physics Circle home Student Login

Home›Chapters›Nuclear Physics PYQs›Law of Radioactive Decay›Question 100

NEET 2012 (AIPMT, Prelims) · Law of Radioactive Decay

Law of Radioactive Decay question from NEET 2012 (AIPMT, Prelims)

Q100NEET 2012 · AIPMT · PRELIMSMedium

Law of Radioactive Decay

A mixture consists of two radioactive materials A1 and A2 with half lives of 20 s and 10 s respectively. Initially the mixture has 40 g of A1 and 160 g of A2. The amount of the two in the mixture will become equal after

Show answer and solution

Answer: (d)

  1. Mass left m=m0(1/2)t/Tm = m_0 (1/2)^{t/T}.
  2. 40⋅2−t/20=160⋅2−t/10⇒2t/10−t/20=4=2240 \cdot 2^{-t/20} = 160 \cdot 2^{-t/10} \Rightarrow 2^{t/10 - t/20} = 4 = 2^2.
  3. t20=2⇒t=40\frac{t}{20} = 2 \Rightarrow t = 40 s.
  4. So the answer is (d) 40 s.
Report a mistake

More questions on Law of Radioactive Decay