The half-life of ${}_{38}^{90}{Sr}$ is 28 years. What is the disintegration rate of 15 mg of this isotope?

Solution

Given,
Half- life of ${}_{38}^{90}{Sr}$ = 28 years.

Using the formula,
$λ = \frac{0.693}{T}$

$\Rightarrow$ $λ = \frac{0.693}{28 \times 365 \times 24 \times 60 \times 60}$

$= 7.85 \times 10^{- 10} s^{- 1}$
90 g of Sr contains 6.023 x 10²³ atoms.

$∴$ 15 mg of Sr contains,

$N_{0} = \frac{6.023 \times 10^{23} \times 15 \times 10^{- 3}}{90} atoms$

$N_{0} = 1.0038 \times 10^{20} atoms$

Disintegration rate, $\frac{dN}{dt} = - λ N_{0}$

$= - 7.85 \times 10^{- 10} \times 1.0038 \times 10^{20} = 7.88 \times 10^{10} dps or Bq = \frac{7.88 \times 10^{10}}{3.7 \times 10^{10}} Ci$

$= 2.13 Ci .$

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The half-life of ${}_{38}^{90}{Sr}$ is 28 years. What is the disintegration rate of 15 mg of this isotope?

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The half-life of Sr3890 is 28 years. What is the disintegration rate of 15 mg of this isotope?

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Mock Test Series

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Entrance Exams Preparation

The half-life of ${}_{38}^{90}{Sr}$ is 28 years. What is the disintegration rate of 15 mg of this isotope?