The decay constant for a radioactive nuclide | vedclass

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Question

$\lambda=1.5 	imes 10^{-5} s ^{-1}$

No. of mole $=\frac{1 	imes 10^{-6}}{60}=\frac{10^{-7}}{6}$

No. of atoms $=$ no. of moles $	imes N _{ A }

Vedclass · Accepted Answer

$15$

Vedclass · Answer

$\lambda=1.5 	imes 10^{-5} s ^{-1}$

No. of mole $=\frac{1 	imes 10^{-6}}{60}=\frac{10^{-7}}{6}$

No. of atoms $=$ no. of moles $	imes N _{ A }$ $=\frac{10^{-7}}{6} 	imes 6 	imes 10^{23}=10^{16}$

$A=N_0 \lambda e^{-\lambda t}$

For, $t =0, A = A _0= N _0 \lambda$

$=1.5 	imes 10^{-5} 	imes 10^{16}=15 	imes 10^{10} Bq .$

The decay constant for a radioactive nuclide is $1.5 \times 10^{-5} s ^{-1}$. Atomic of the substance is $60\,g$ mole $^{-1},\left( N _{ A }=6 \times 10^{23}\right)$. The activity of $1.0\,\mu g$ of the substance is $.......\,\times 10^{10}\,Bq$

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