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(^{238}U) = 238.050784 u, m(^{95}Sr) ...

For more information, see **Fission** | **Physics**

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 ...

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 ...

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.

For more information, see World **Energy** Resources | 2016

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