A) 4.8 yr
B) 7 yr
C) 10 yr
D) 9.6 yr
Correct Answer: C
Solution :
Number of atoms remained undecayed after time t, \[N={{N}_{0}}{{e}^{-\lambda t}}\] Given, \[N=\frac{3{{N}_{0}}}{100}\] \[\therefore \] \[\frac{3{{N}_{0}}}{100}={{N}_{0}}{{e}^{-\lambda t}}\] or \[3=100{{e}^{-\lambda t}}\] ... (i) Now, \[\lambda =\frac{0.693}{{{T}_{1/2}}}=\frac{0.693}{2}=0.346\] Substituting the value of \[\lambda \] in Eq. (i), we get \[2.303\,\log \,\left( \frac{3}{100} \right)=-0.346\,t\] \[-1.52\times 2.303=-0.346\,t\] \[\Rightarrow \] \[t=10\,yr\]
You need to login to perform this action.
You will be redirected in
3 sec
