Question

Calculate the net amount of energy released in the deuterium-tritium reaction,${}_1{}^{2}D+{}_1{}^{3}T\to {}_2{}^{4}He+{}_0{}^{1}n$.

Short answer

$Q=17.45MeV$

Step-by-step solution

Given data and concept

Reaction$D_1^2+T_1^3\to H{e}_2^4+n_0^1$

• ${\text{m}}_{{\text{D}}_{\text{1}}^{\text{2}}}\text{= 2.014u}$- the mass of deuterium
• $m_{T_1^3}=3.016u$- the mass of tritium
• ${\text{m}}_{{\text{He}}_{\text{2}}^{\text{4}}}\text{= 4.00260u}$- the mass of helium
• ${\text{m}}_{{\text{n}}_{\text{0}}^{\text{1}}}\text{= 1.008665u}$- the mass of the neutron

We will be applying an equation that determines energy as:

role="math" localid="1658422170797" $Q=\Delta m{c}^2$

Where:$\Delta m$ - mass defect,C - speed of light

Where the mass defect$\Delta m$ is given by:$\Delta m=m_{D_1^2}+m_{T_1^3}-m_{H{e}_2^4}-m_{n_0^1}{c}^2$

Put the known values in the equation

Calculating the Q for the reaction:

$\begin{array}{rcl}Q& =& (2.0194+3.0164-4.00260-1.008665)u{c}^2\\ & =& 0.0187354\times 931.5MeV\\ & =& 17.45MeV\\ & & \end{array}$

Therefore, the energy released is $Q=17.45MeV$.