16:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 17:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 15:T50a,{"@context":"https://schema.org","@type":"Quiz","name":"Mht Cet 2026 Atoms And Nuclei PYQ with Solution","url":"https://pyq-test-series-platform-alsjdcxu2-sameers-projects-41c5a71e.vercel.app/mht-cet/questions/2026/physics/atoms-and-nuclei/jt-mr4o54o8","educationalLevel":"Entrance exam preparation","about":"Mht Cet / Physics / Atoms And Nuclei","hasPart":{"@type":"Question","name":"Mht Cet 2026 Atoms And Nuclei PYQ with Solution","text":"The activity of a radioactive sample is measured as N_0 counts per minute at time t = 0 and \\dfrac{N_0}{e} counts per minute at time t = 3 minute. The time (in minute) in which the activity reduces to half the value, is","eduQuestionType":"mcq","acceptedAnswer":{"@type":"Answer","text":"B"},"comment":{"@type":"Comment","text":"The activity of a radioactive sample decreases exponentially with time. So we use N = N_0 e^{-\\lambda t} where \\lambda is the decay constant. Given that at t = 3 minute, N = \\frac{N_0}{e} Substitute in the formula: \\frac{N_0}{e} = N_0 e^{-3\\lambda} Cancel N_0 from both sides: \\frac{1}{e} = e^{-3\\lambda} We know \\frac{1}{e} = e^{-1} So, e^{-3\\lambda} = e^{-1} Hence, -3\\lambda = -1 \\lambda = \\frac{1}{3} Now, half-life is the time in which activity becomes half of its initial value. For half-life,..."}}}18:T470,

The activity of a radioactive sample decreases exponentially with time.

So we use

$$ N = N_0 e^{-\lambda t} $$

where $\lambda$ is the decay constant.

Given that at $t = 3$ minute,

$$ N = \frac{N_0}{e} $$

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