Physics Circle home Student Login

Home›Chapters›Nuclear Physics PYQs›Law of Radioactive Decay›Question 77

JEE Main 2020 (5 Sep, Shift 1) · Law of Radioactive Decay

Law of Radioactive Decay question from JEE Main 2020 (5 Sep, Shift 1)

Q77JEE Main 2020 · 5 SEP · SHIFT 1Medium

Law of Radioactive Decay

Activities of three radioactive substances A, B and C are represented by the curves A, B and C, in the figure. Then their half-lives

T12(A)T_{\frac{1}{2}}(A): T12(B)T_{\frac{1}{2}}(B): T12(C)T_{\frac{1}{2}}(C) are in the ratio:

Figure

Show answer and solution

Answer: (d)

  1. R=R0e−λt⇒ln⁡R=ln⁡R0−λtR = R_0 e^{-\lambda t} \Rightarrow \ln R = \ln R_0 - \lambda t, so λ\lambda = magnitude of slope.
  2. λA=6/10=0.6\lambda_A = 6/10 = 0.6, λB=6/5=1.2\lambda_B = 6/5 = 1.2, λC=2/5=0.4\lambda_C = 2/5 = 0.4 per year.
  3. T1/2∝1λT_{1/2} \propto \frac{1}{\lambda}: 10.6:11.2:10.4=2:1:3\frac{1}{0.6} : \frac{1}{1.2} : \frac{1}{0.4} = 2 : 1 : 3.
  4. So the answer is (d) 2 : 1 : 3.
Report a mistake

More questions on Law of Radioactive Decay