question_answer

(i) Some scientist have predicted that a global nuclear war on the earth would be followed by nuclear winter. What would cause nuclear winter?

(ii) Calculate the activity of 1 mg sample of \[{}_{37}Sr{}^{90},\] whose half-life period is 28 yr,

(iii) Define decay constant.

Or

Suppose India has a target of producing by 2020 AD 200,000 MW of electric power, 10 per cent of which is to be obtained from nuclear power plants. Suppose we are given that, on average the efficiency of utilisation (i.e., conversion to electric energy) of thermal energy produced in a reactor is 25%. How much amount of fissionable uranium would our country need per year at the turn of this century?

Take the heat energy per fission of \[{{U}^{235}}\] to be about 200 MeV Avogadro's number \[{{N}_{A}}=6.023\times {{10}^{23}}mo{{l}^{-1}}.\]

Answer:

(ii) \[5.24\times {{10}^{9}}decay/s\] Or \[3.078\text{ }x{{10}^{4}}\,kg\] (i) The clouds produced by a global nuclear war would perhaps cover major parts of the sky preventing solar light from reaching many parts of the earth. This would cause a ?nuclear winter?. (ii) Here, number of atoms in 90 g of \[_{37}S{{r}^{90}}\]             \[=6.023\times {{10}^{23}}\] Number of atoms in 1 mg of \[_{37}S{{r}^{90}}\]             \[=\frac{6.023\times {{10}^{23}}}{90}\times \frac{1}{1000}\]             \[=0.669\times {{10}^{19}}\]             \[{{T}_{1/2}}28\times 3.156\times {{10}^{7}}s\] \[\therefore \]Activity of sample,\[R=\lambda N\left( \because \lambda =\frac{0.693}{{{t}_{1/2}}} \right)\]             \[=\frac{0.693}{{{T}_{1/2}}}N=\frac{0.693\times 0.669\times {{10}^{19}}}{28\times 3.156\times {{10}^{7}}}\]             \[=0.005246\times {{10}^{12}}=5.2\times {{10}^{9}}Bq\] (iii) Decay constant It may be defined as the reciprocal of the time interval during which the number of active nuclei in a given substance (radioactive) reduced to 36.8% (or\[\frac{1}{e}\]times) of its initial value. Or Target of power by 2020 AD             \[=2\times {{10}^{5}}MW=2\times {{10}^{11}}W\] Power required from nuclear power plants             \[=10%\,\,\text{of}\,\,2\times {{10}^{11}}W\]             \[=\frac{10}{100}\times 2\times {{10}^{11}}=2\times {{10}^{10}}W\] \[\therefore \]Energy required from nuclear power plants per year             \[=\text{power}\times \text{time}\]             \[=2\times {{10}^{10}}\times 365.25\times 24\times 60\times 60\,\,\text{J}\]             \[=6.312\times {{10}^{17}}\,\,\text{J}\] Energy released per fission = 200 MeV Available electrical energy per fission of\[{{U}^{235}}\]nucleus             \[=25%\,\,of\,\,200MeV=\frac{25}{100}\times 200MeV\]             \[=25\times 2\times 1.6\times {{10}^{-19}}\times {{10}^{6}}\text{J}\]             \[=8\times {{10}^{-12}}\text{J}\]                         \[[\because 1\,\,MeV=1.6\times {{10}^{-13}}J]\] Number of \[{{U}^{235}}\] fissions required per year             \[=\frac{6.312\times {{10}^{17}}}{8\times {{10}^{-12}}}=7.89\times {{10}^{28}}\] Required number of moles of \[{{U}^{235}}\]             \[=\frac{7.89\times {{10}^{28}}}{\text{Avogadro }\!\!'\!\!\text{ s}\,\,\text{number}}\]             \[=\frac{7.89\times {{10}^{28}}}{6.028\times {{10}^{23}}}\]             \[=1.31\times {{10}^{5}}\]moles Mass of \[{{U}^{235}}\] required = number of moles mass \[\times \]number             \[=1.31\times {{10}^{5}}\times 235g\]             \[=307.85\times {{10}^{5}}kg\]             \[=3.0785\times {{10}^{7}}kg\]