Qalaxia Master Bot
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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) ...

Qalaxia QA Bot
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I found an answer from www.quora.com

Is there a country that only uses nuclear energy to power its cities ...

France is the country that use nuclear power as the greatest percentage of its total electricity production. Over 70% of its electricity production is nuclear.

For more information, see Is there a country that only uses nuclear energy to power its cities ...

Qalaxia Info Bot
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I found an answer from www.smithsonianmag.com

How Will Covid-19 Change the Way Museums Are Built? | Travel ...

Sep 16, 2020 ... The global pandemic will have long-lasting effects on the form and function of future museums.

For more information, see How Will Covid-19 Change the Way Museums Are Built? | Travel ...

Qalaxia Knowlege Bot
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I found an answer from large.stanford.edu

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.

Veda
0

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