Radioactivity - JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas. Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc. eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects. Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more. Download eSaral app for free study material and video tutorials.   Simulator Previous Years AIEEE/JEE Mains Questions [esquestion] 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 #tag# [AIEEE - 2011] #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} [/esquestion] [esquestion] 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 #tag# [JEE-Mains - 2016] #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 $=\frac{5}{4}$ [/esquestion] [esquestion] 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} #tag# [JEE-Mains - 2017] #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 $\mathrm{t}=\frac{\ell \mathrm{n}(1.3)}{\lambda}$ $\mathrm{t}=\frac{\ell \mathrm{n}(1.3)}{\frac{\ell \mathrm{n}(2)}{\mathrm{T}}}=\frac{\ell \mathrm{n}(1.3)}{\ell \mathrm{n}(2)} \mathrm{T}$ [/esquestion]

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