Starting with a sample of pure Cu 66 , 7 8 of it decays into Zn in 15 minutes. The corresponding half-life is

The undecayed mass left after time t is given by N = N 0 1 2 t T Here, t = 15 min ; N = N 0 - 7 N 0 8 = N 0 8 ∴ N 0 8 = N 0 1 2 15 T Or 2 15 T = 8 = 2 3 Or T = 15 3 = 5 min

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PhysicsModern PhysicsRadioactivityEasy2 minPYQ_2020

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Starting with a sample of pure ${Cu}^{66}, \frac{7}{8}$ of it decays into $Zn$ in $15$ minutes. The corresponding half-life is

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Solution:

The undecayed mass left after time $t$ is given by

$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T}}$

Here, $t = 15 \min; N = N_{0} - \frac{7 N_{0}}{8} = \frac{N_{0}}{8}$

$∴$ $\frac{N_{0}}{8} = N_{0} {(\frac{1}{2})}^{\frac{15}{T}}$

Or $2^{\frac{15}{T}} = 8 = 2^{3}$

Or $T = \frac{15}{3} = 5 \min$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Radioactivity

⏱ 2mℹ️ Source: PYQ_2020

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