energy released per fusion - Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of ^{235} U to be about 200 MeV.

Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of ^{235} U to be about 200 MeV.

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Anonym0us (Student, UG1 Grade)

Qalaxia Master Bot (last edited 1 month ago)

I found an answer from courses.lumenlearning.com

Fission | Physics

Fission is the opposite of fusion and releases energy only when heavy nuclei are split. ... The amount of energy per fission reaction can be large, even by nuclear standards. ... about 1 MeV per nucleon, or approximately 240 MeV per fission, is released. ... given the atomic masses to be m(²³⁸U) = 238.050784 u, m(⁹⁵Sr) ...

For more information, see Fission | Physics

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World Energy Resources | 2016

Oct 13, 2016 ... Natural gas is the second largest energy source in power generation, ... percentage of renewable energy in electricity production in the various regions. ... solutions for supply (i.e. renewables) and demand (i.e. enhancing energy efficiency) and ... In terms of direct use of geothermal heat, the countries with.

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Veda (last edited 1 month ago) (Student, UG2 Grade)

Given that

The amount of electricity that will be generated E = 200000 MW

Nuclear power plants are needed to provide the energy = 10% of E

= \frac{10}{100} * 200000 = 20000 MW

= 20000 * 10^{6} joules/sec

Nuclear power per year (Converting joules/sec to joules/year)

= 2 * 10^{10} * 365*24*60*60 = 6.307*10^{17} joules/year

The energy emitted by a single uranium nucleus fission is 200 MeV.

Per fusion, the amount of nuclear energy converted into electrical energy = 25 % of 200 MeV

\frac{25}{100} * 200 = 50 MeV = 50 * 10^{6} * 1.6* 10^{-19} = 8* 10^{-12} Joules

Step 1: Determining the number of atoms required to per fission per year

[math]

number of fissions per year = \frac{6.307*10^{17}}{8* 10^{-12}} = 78840* 10^{24} atoms

Step 2: Finding the mass of nuclear substance per year

1 g (1 mole) of the uranium contains 6.023* 10^{23} atoms

Atomic mass of uranium = 235 g = 235* 10^{-3} Kg

Mass of 78840* 10^{24} uranium atoms = \frac{235* 10^{-3}}{6.023* 10^{23}} * 78840* 10^{24}

= 3.016 * 10^{4} kg

As a results, the amount of uranium required each year = 3.016 * 10^{4} kg