JEE Main 2023PhysicsNuclear PhysicsMediumNumerical

JEE Main 2023 — Nuclear Physics Question with Solution

JEE Main 2023 (10 Apr Shift 1)

Question

The decay constant for a radioactive nuclide is 1.5×10-5 s-1. Atomic weight of the substance is 60 g mole-1, NA=6×1023. The activity of 1.0 µg of the substance is ________ ×1010 Bq.

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Show full solutionCorrect answer: 15
Correct answer
15

Step-by-step explanation

It is given that 

λ=1.5×10-5 s-1

Number of moles is 1×10-660=10-76

Number of atoms=number of moles ×NA

N0=10-76×6×1023=1016

Hence, at t=0, A=A0=N0λ, the activity can be given by

A=N0λe-λt⇒A=1016×1.5×10-5×1=15×1010 Bq

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About this question

This is a previous-year question from JEE Main 2023, covering the Nuclear Physics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.