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13.Nuclei
hard
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 $.......\,\times 10^{10}\,Bq$
A
$14$
B
$13$
C
$12$
D
$15$
(JEE MAIN-2023)
Solution
$\lambda=1.5 \times 10^{-5} s ^{-1}$
No. of mole $=\frac{1 \times 10^{-6}}{60}=\frac{10^{-7}}{6}$
No. of atoms $=$ no. of moles $\times N _{ A }$ $=\frac{10^{-7}}{6} \times 6 \times 10^{23}=10^{16}$
$A=N_0 \lambda e^{-\lambda t}$
For, $t =0, A = A _0= N _0 \lambda$
$=1.5 \times 10^{-5} \times 10^{16}=15 \times 10^{10} Bq .$
Standard 12
Physics
