The decay constant for a radioactive nuclide is 1 . 5 × 10 ^ - 5 s...

It is given that

$\lambda = 1 . 5 \times {10}^{- 5}{s}^{- 1}$

Number of moles is $\dfrac{1 \times {10}^{- 6}}{60} = \dfrac{{10}^{- 7}}{6}$

Number of atoms=number of moles $\times {N}_{A}$

${N}_{0} = \dfrac{{10}^{- 7}}{6} \times 6 \times {10}^{23} = {10}^{16}$

Hence, at $t = 0, A = {A}_{0} = {N}_{0}\lambda$, the activity can be given by

$A = {N}_{0}\lambda{e}^{- \lambdat} ⇒ A = {10}^{16} \times 1 . 5 \times {10}^{- 5} \times 1 = 15 \times {10}^{10}Bq$