A sample of radioactive material contains initially 3 × 1 0 18 atoms. The material decays by α − emission and has a half life of 5 × 1 0 7 sec . The initial number of disintegration per second will be approximately.

d t d N = λ N 0 = 5 × 1 0 7 s ln 2 × 3 × 1 0 18 d t d N ≈ 4.2 × 1 0 10 disintegrations per second Hence option (C).

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A sample of radioactive material contains initially $3 \times 1 0^{18}$ atoms. The material decays by $α -$ emission and has a half life of $5 \times 1 0^{7} sec$ . The initial number of disintegration per second will be approximately.

Options:

Answer:

Solution:

$\frac{d N}{d t} = λ N_{0} = \frac{ln 2}{5 \times 1 0 ^{7} s} \times 3 \times 1 0^{18}$

$\frac{d N}{d t} \approx 4.2 \times 1 0^{10} disintegrations per second$

Hence option (C).

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Radioactivity

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