calculate the energy released in the nuclear fusion of isotopes of hydrogen
(i) `._(1)^(2)H + ._(1)^(2)H rarr ._(2)^(3)He + ._(0)^(1)n`
(ii) `._(1)^(2)H + ._(1)^(3)H rarr ._(2)^(4)He + ._(0)^(1)n`
Given that mass of neutron = 1.00867 amu
`{:("mass of",._(1)^(2)H = 2.01410,am u,),("mas of",._(1)^(3)H = 3.01603,am u,),("mass of ",._(2)^(3)H = 3.0160,am u,),("mass of ",._(2)^(4)He = 4.00260,am u,):}`

Answer

To calculate the energy released in the nuclear fusion of isotopes of hydrogen, we will follow these steps for both reactions.

### Given Data:
- Mass of neutron, \( m_n = 1.00867 \, \text{amu} \)
- Mass of deuterium, \( m_{D} = 2.01410 \, \text{amu} \)
- Mass of tritium, \( m_{T} = 3.01603 \, \text{amu} \)
- Mass of helium-4, \( m_{He} = 4.00260 \, \text{amu} \)

### (i) Reaction:
\[ _{1}^{2}H + _{1}^{2}H \rightarrow _{2}^{3}He + _{0}^{1}n \]

1. **Calculate the total mass of reactants:**
\[ \text{Total mass of reactants} = 2 \times m_{D} = 2 \times 2.01410 = 4.02820 \, \text{amu} \]

2. **Calculate the total mass of products:**
\[ \text{Total mass of products} = m_{He} + m_n = 3.0160 + 1.00867 = 4.02467 \, \text{amu} \]

3. **Calculate the mass defect (\( \Delta m \)):**
\[ \Delta m = \text{Total mass of reactants} - \text{Total mass of products} = 4.02820 - 4.02467 = 0.00353 \, \text{amu} \]

4. **Convert mass defect to energy using \( E = \Delta m \cdot c^2 \):**
\[ E = 0.00353 \, \text{amu} \times 931 \, \text{MeV/amu} = 3.28 \, \text{MeV} \]

### (ii) Reaction:
\[ _{1}^{2}H + _{1}^{3}H \rightarrow _{2}^{4}He + _{0}^{1}n \]

1. **Calculate the total mass of reactants:**
\[ \text{Total mass of reactants} = m_{D} + m_{T} = 2.01410 + 3.01603 = 5.03013 \, \text{amu} \]

2. **Calculate the total mass of products:**
\[ \text{Total mass of products} = m_{He} + m_n = 4.00260 + 1.00867 = 5.01127 \, \text{amu} \]

3. **Calculate the mass defect (\( \Delta m \)):**
\[ \Delta m = \text{Total mass of reactants} - \text{Total mass of products} = 5.03013 - 5.01127 = 0.01886 \, \text{amu} \]

4. **Convert mass defect to energy using \( E = \Delta m \cdot c^2 \):**
\[ E = 0.01886 \, \text{amu} \times 931 \, \text{MeV/amu} = 17.55 \, \text{MeV} \]

### Final Answers:
- Energy released in reaction (i): \( 3.28 \, \text{MeV} \)
- Energy released in reaction (ii): \( 17.55 \, \text{MeV} \)