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PhysicsModern PhysicsRadioactivityEasy2 minPYQ_2024

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A radioactive substance has half-life of 60 minute. During 3 hour, the amount of substance decayed would be

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Solution:

Given half-life $T_{1/2} = 60 minute = 1 hour$ .

Time elapsed $t = 3 hour$ .

Number of half-lives $n = \frac{t}{T _{1/2}} = \frac{3}{1} = 3$ .

Amount of substance remaining after $n$ half-lives is given by $N = N_{0} (\frac{1}{2})^{n}$ .

Substituting $n = 3$ , we get $N = N_{0} (\frac{1}{2})^{3} = \frac{N _{0}}{8}$ .

Amount of substance decayed is $N_{0} - N = N_{0} - \frac{N _{0}}{8} = \frac{7 N _{0}}{8}$ .

Percentage of substance decayed is $\frac{7}{8} \times 100% = 87.5%$ .

Answer: $87.5%$

Stream:MHT_CETSubject:PhysicsTopic:Modern PhysicsSubtopic:Radioactivity

⏱ 2mℹ️ Source: PYQ_2024

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