A) \[20\mu g\]
B) \[40\mu g\]
C) \[1\,\mu g\]
D) \[10\,\mu g\]
Correct Answer: B
Solution :
| Let assume power \[P=1000\text{ }W\] |
| Energy per hour \[=1000\times 3600\text{ }J\] |
| Energy per fission \[=200\text{ }MeV\] |
| \[=200\times 1.6\times {{10}^{-13}}J\] |
| \[\therefore \] Number of fission per hour |
| \[n=\frac{1000\times 3600}{200\times 1.6\times {{10}^{-13}}}\] |
| Number of mole per hour \[=\frac{n}{N}\times 235\] |
| \[\therefore \] Mass per hour \[=\frac{n}{N}\times 235\] |
| \[=\frac{1000\times 3600\times 235}{200\times 1.6\times {{10}^{-13}}\times 6.02\times {{10}^{23}}}\] |
| \[=43.9\times {{10}^{-6}}g\] |
| This \[43.9\times {{10}^{-6}}g\] is nearest value of 40 micron so option [b] is correct. |
You need to login to perform this action.
You will be redirected in
3 sec
