A) 6400 years
B) 4800 years
C) 3200 years
D) 1600 years
Correct Answer: B
Solution :
For the relation \[N={{N}_{0}}{{e}^{-\lambda t}}\] where \[\lambda =\frac{0.693}{{{T}_{1/2}}}\] Hence \[{{T}_{1/2}}=1600\,\] years, \[{{N}_{0}}=400g,\] \[N=400-100=300g\] so,\[-\frac{0.693t}{{{T}_{1/2}}}=\frac{300}{400}\] or \[\frac{0.693}{{{T}_{1/2}}}t=\ln \frac{3}{4}\] so \[t=\frac{{{T}_{1/2}}\ln \frac{3}{4}}{0.693}=\frac{1600\,\ln \frac{3}{4}}{0.693}\] \[=4800\text{ }years\]You need to login to perform this action.
You will be redirected in
3 sec