A) \[{{10}^{-23}}g\]
B) \[{{10}^{-15}}g\]
C) \[{{10}^{-16}}g\]
D) \[{{10}^{-14}}g\]
Correct Answer: B
Solution :
We know that 1 Curie \[3.7\,\,\times \,\,{{10}^{10}}\] disintegration per sec So, activity \[=6.023\times 3.7\times {{10}^{10}}\,\text{dps}\] \[\lambda =3.7\times {{10}^{14}}\,{{\sec }^{-1}}\] \[N=\frac{Activity}{\lambda }=\frac{6.023\times 3.7\times {{10}^{-10}}}{3.7\times {{10}^{4}}}\] \[=6.023\times {{10}^{6}}\text{atoms}\] \[\because \] Weight of \[6.023\times {{10}^{23}}\] atoms = 100 g \[\therefore \] Weight of \[6.023\times {{10}^{6}}\] atoms will be \[=\frac{100\times 6.023\times {{10}^{6}}}{6.023\times {{10}^{23}}}={{10}^{-15}}\,g\]
You need to login to perform this action.
You will be redirected in
3 sec
