Question

How much energy will approximately be  released if all the atoms of 1 kg of deuterium  could underg fusion? 

1. 2 × 10⁷ k Wh
2. 8 × 10¹³ J
3. 6 × 10²⁷ colorie
4. 8 × 10²³ MeV

Answer 1

Correct Answer - Option 2 : 8 × 10¹³ J

Concept:

• Deuterium (2 H 1): Deuterium is an isotope of hydrogen having mass number 2 and atomic number 1.
• Nuclear Fusion Reaction: The reaction in which two small nuclei combine to form a bigger nucleus. Huge energy is released in the nuclear fusion reaction.
• Example of Nuclear Fusion Reaction: Two Deuterium atoms combine to form one Helium Atom. 

1 H 2 + 1 H 2 → 2 He 4 + 3.27 MeV

• This amount of energy (3. 27 MeV)  is generated when 2 deuterium atom each having 2 unit of atomic mass is reacting.
• Electron Volt (eV): Electron volt is a unit of energy. 1 Electron volt is the energy acquired by 1 electron when it is accelerated by the potential difference of 1 volt. 

1eV = 1.602 × 10 ^-19 Joule

• Mole Concept: The mass present in 6. 023 × 10 ²³ atoms of an element is equal to the numerical atomic mass of the substance expressed in grams.
• Example: Mass of deuterium atom is 2 atomic mass unit. In 2 gram of Deuterium 6. 023 × 10 23 atoms are present.

Calculation:

In 2 gram of deuterium,  6. 023 × 10 23 atoms are present

In 1000 gram (1 kg) of deuterium atom, \(\frac{6.023 × 10^{23}}{2}× 1000\) atoms are present 

In 500 gram (0.5 kg) of deuterium atom, \(\frac{6.023 × 10^{23}}{4}× 1000\) atoms are present 

For a fusion reaction, two deuterium atoms are required. In 1 kg of deuterium, 500 grams of deuterium will react with other 500 grams.

Since energy released in the fusion of 1 deuterium reacting with another is 3. 27 eV

The energy released in the fusion of 500 gm(\((\frac{6.023 × 10^{23}}{4}× 1000) atoms\)) atoms of deuterium 

\(E = (\frac{6.023 × 10^{23}}{4}× 1000) × 3.27 MeV\)

1 eV = 1.602 × 10 -19 Joule

1 MeV = 1.602 × 10 -19 × 10 ⁶ Joule = 1.602 × 10 -13 Joule

3.27 MeV = 3.27 × 1.602 × 10 -13 Joule = 5,23 × 10 -13 Joule

Now, total energy required

\(E = (\frac{6.023 × 10^{23}}{4}× 1000) × 5.23 × 10^{-13}\)

E = 7.87 × 10 ¹³ Joule

⇒ E = 7.87 × 10 13 Joule ≈ 8 × 10 13 Joule

So, 8 × 10 13 Joule is the correct answer.

• Nuclear fission is another type of nuclear reaction in which one heavier nucleus breaks into two or more nuclei. This mostly used in nuclear power plants.
• Nuclear fusion reactions generate far more power than fission. It is difficult to control, hence not much-used practice for energy generation.
• Nuclear fusion reactions are continually taking place on Sun.