The decay constant for a radioactive nuclide is . Atomic of the substance is , . The activity of of the substance is _________ . [2023]

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Solution

Q.
The decay constant for a radioactive nuclide is $1.5 \times 10^{- 5} s^{- 1}$ . Atomic of the substance is $60 {g mole}^{- 1}$ , $(N_{A} = 6 \times 10^{23})$ . The activity of $1.0 μ g$ of the substance is _________ $\times 10^{10} Bq$ . [2023]

Ans.
(15)

$λ = 1.5 \times 10^{- 5} s^{- 1}$

$Number of mole = \frac{1 \times 10^{- 6}}{60} = \frac{10^{- 7}}{6}$

$No. of atoms = no. of moles \times N_{A}$

$= \frac{10^{- 7}}{6} \times 6 \times 10^{23} = 10^{16}$

$A = N λ e^{- λ t}$

$For t = 0, A_{0} = N_{0} λ = 1.5 \times 10^{- 5} \times 10^{16} = 15 \times 10^{10} Bq$

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