Radioactivity – JEE Main Previous Year Questions with Solutions
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Q. The half life of a radioactive substance is 20 minutes. The approximate time interval $\left(\mathfrak{t}_{2}-\mathfrak{t}_{1}\right)$ between the time $t_{2}$ when $\frac{2}{3}$ of it has decayed and time $t_{1}$ when $\frac{1}{3}$ of it had decayed is :- (1) 20 min (2) 28 min (3) 7 min (4) 14 min [AIEEE – 2011]

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Sol. (1) \because \frac{\mathrm{N}}{\mathrm{N}_{0}}=\left[\frac{1}{2}\right]^{t / T} \quad \therefore \frac{1}{3}=\left[\frac{1}{2}\right]^{\frac{t_{2}}{T}} \& \frac{2}{3}=\left[\frac{1}{2}\right]^{\frac{t_{1}}{T}} \Rightarrow \frac{1}{2}=\left[\frac{1}{2}\right]^{\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right) \frac{1}{\mathrm{T}}} \Rightarrow 1=\frac{\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)}{\mathrm{T}} \Rightarrow \mathrm{T}=\mathrm{t}_{2}-\mathrm{t}_{1} \Rightarrow \mathrm{t}_{2}-\mathrm{t}_{1}=20 \mathrm{min}

Q. Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the samples have equal number of nuclei. After 80 minutes , the ratio of decayed numbers of A and B nuclei will be :- (1) 5 : 4 (2) 1 : 16 (3) 4 : 1 (4) 1 : 4 [JEE-Mains – 2016]

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Sol. (1) \mathrm{t}=80 \mathrm{min}=4 \mathrm{T}_{\mathrm{A}}=2 \mathrm{T}_{\mathrm{B}} \therefore \text { no. of nuclei of } \mathrm{A} \text { decayed }=\mathrm{N}_{0}-\frac{\mathrm{N}_{0}}{2^{4}}=\frac{15 \mathrm{N}_{0}}{16} \therefore \text { no. of nuclei of } \mathrm{B} \text { decayed }=\mathrm{N}_{0}-\frac{\mathrm{N}_{0}}{2^{2}}=\frac{3 \mathrm{N}_{0}}{4} required ratio $=\frac{5}{4}$

Q. A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by : (1) $\mathrm{t}=\mathrm{T} \log (1.3)$ $(2) \mathrm{t}=\frac{\mathrm{T}}{\log (1.3)}$ (3) $\mathrm{t}=\frac{\mathrm{T}}{2} \frac{\log 2}{\log 1.3}$ (4) \mathrm{t}=\mathrm{T} \frac{\log 1.3}{\log 2} [JEE-Mains – 2017]

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Sol. (4) At time t $\frac{\mathrm{N}_{\mathrm{B}}}{\mathrm{N}_{\mathrm{A}}}=.3 \Rightarrow \mathrm{N}_{\mathrm{B}}=.3 \mathrm{N}_{\mathrm{A}}$ also let initially there are total $\mathrm{N}_{0}$ number of nuclei $\mathrm{N}_{\mathrm{A}}+\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{0}$ $\mathrm{N}_{\mathrm{A}}=\frac{\mathrm{N}_{0}}{1.3}$ Also as we know $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{N}_{0}}{1.3}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{1}{1.3}=\mathrm{e}^{-\lambda t} \Rightarrow \ell \mathrm{n}(1.3)=\lambda \mathrm{t}$ or $\mathrm{t}=\frac{\ell \mathrm{n}(1.3)}{\lambda}$ $\mathrm{t}=\frac{\ell \mathrm{n}(1.3)}{\frac{\ell \mathrm{n}(2)}{\mathrm{T}}}=\frac{\ell \mathrm{n}(1.3)}{\ell \mathrm{n}(2)} \mathrm{T}$