A) 4800 year
B) 6400 year
C) 2400 year
D) 3200 year
Correct Answer: D
Solution :
| Amount of substance remained is |
| \[M={{M}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] |
| Given, |
| \[{{M}_{0}}=100\,g;\,\,M=25\,kgm,\,\,{{T}_{1/2}}=1600\] years |
| So, \[25=100{{\left( \frac{1}{2} \right)}^{n}}\] |
| or \[\frac{25}{100}={{\left( \frac{1}{2} \right)}^{n}}\] |
| or \[{{\left( \frac{1}{2} \right)}^{2}}={{\left( \frac{1}{2} \right)}^{n}}\] |
| Comparing the power, we have |
| \[n=2\] |
| or \[\frac{t}{{{T}_{1/2}}}=2\] |
| or \[t=2{{T}_{1/2}}=2\times 1600=3200\] years |
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