Chapter 11: Q62E (page 520)

Calculate the net amount of energy released in the deuterium-tritium reaction, ${}_{1}^{2}D + {}_{1}^{3}T \to {}_{2}^{4}H e + {}_{0}^{1}n$ .

Short Answer

Expert verified

$Q = 17.45 M e V$

Step by step solution

Given data and concept

Reaction $D_{1}^{2} + T_{1}^{3} \to H e_{2}^{4} + n_{0}^{1}$

$m_{D_{1}^{2}} = 2.014u$ - the mass of deuterium
$m_{T_{1}^{3}} = 3.016 u$ - the mass of tritium
$m_{{He}_{2}^{4}} = 4.00260u$ - the mass of helium
$m_{n_{0}^{1}} = 1.008665u$ - the mass of the neutron

We will be applying an equation that determines energy as:

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

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

Where the mass defect $Δ m$ is given by: $Δ m = m_{D_{1}^{2}} + m_{T_{1}^{3}} - m_{H e_{2}^{4}} - m_{n_{0}^{1}} c^{2}$

Put the known values in the equation

Calculating the Q for the reaction:

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

Therefore, the energy released is $Q = 17.45 M e V$ .

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Calculate the net amount of energy released in the deuterium-tritium reaction, ${}_{1}^{2}D + {}_{1}^{3}T \to {}_{2}^{4}H e + {}_{0}^{1}n$ .

Short Answer

Step by step solution

Given data and concept

Put the known values in the equation

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Step by step solution

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Most popular questions from this chapter

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Calculate the net amount of energy released in the deuterium-tritium reaction, ${}_{1}^{2}D + {}_{1}^{3}T \to {}_{2}^{4}H e + {}_{0}^{1}n$ .