A) \[\text{3}.\text{61}\times \text{l}{{0}^{\text{1}0}}\]
B) \[\text{3}.\text{61}\times \text{l}{{0}^{\text{12}}}\]
C) \[\text{3}.\text{11}\times \text{l}{{0}^{\text{15}}}\]
D) \[\text{31}.\text{1}\times \text{l}{{0}^{\text{15}}}\]
Correct Answer: A
Solution :
From Rutherford Soddy law, the number of atoms that will decay is \[\frac{dN}{dt}=\lambda \,N\] ?. (i) Also, \[\lambda =\frac{0.693}{{{T}_{1/2}}}\] .... (ii) where \[{{T}_{1/2}}\] is half-life of radioactive sample. Given, \[{{T}_{1/2}}=1620\,yr\] ^1620^ \[=1620\times 365\times 24\times 60\times 60\] \[\therefore \] \[\lambda =\frac{0.693}{1620\times 365\times 24\times 60\times 60}\] and \[N=\frac{6.023\times {{10}^{23}}}{226}\] Putting these values of N and \[\lambda \] in Eq. (i), we get \[\frac{dN}{dt}=\frac{0.693\times 6.023\times {{10}^{23}}}{1620\times 365\times 24\times 60\times 60\times 226}\] \[\frac{dN}{dt}=3.61\times {{10}^{10}}\]
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