A)
B)
C)
D)
Correct Answer: D
Solution :
| [d] A time t |
| \[\frac{{{N}_{B}}}{{{N}_{A}}}=.3\Rightarrow {{N}_{B}}=.3{{N}_{A}}\] |
| also let initially there are total N0 number of nuclei |
| \[{{N}_{A}}+{{N}_{B}}={{N}_{0}}\] |
| \[{{N}_{A}}=\frac{{{N}_{0}}}{1.3}\] |
| Also as we know |
| \[{{N}_{A}}={{N}_{0}}{{e}^{-\lambda t}}\] |
| \[\frac{{{N}_{0}}}{1.3}={{N}_{0}}{{e}^{-\lambda t}}\] |
| \[\frac{1}{1.3}={{e}^{-\lambda t}}\ell n(1.3)=\lambda t\]or \[t=\frac{\ell n(1.3)}{\lambda }\] |
| \[t=\frac{\ell n(1.3)}{\frac{\ell n(2)}{T}}=\frac{\ell n(1.3)}{\ell n(2)}T\] |
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