Chapter 11: Q61E (page 520)

Determine Q for the reaction ${}_{3}^{7}L i + {}_{1}^{1}H \to {}_{4}^{7}B e + {}_{0}^{1}n$

Short Answer

Expert verified

Energy released $Q = - 1.64 M e V$

Step by step solution

Given data and concept

The reaction between lithium and hydrogen:

$L i_{3}^{7} + H_{1}^{1} \to B e_{4}^{7} + n_{0}^{1}$

$m_{L_{3}^{7}} = 6.941 u$ - the mass of lithium
$m_{H_{1}^{1}} = 1.0079 u$ - the mass of hydrogen
$m_{B e_{4}^{7}} = 7.01692 u$ - the mass of beryllium
$m_{n_{0}^{2}} = 1.00866 u$ - unknown

So, we had to determine the released heat Q for the reaction.

We will be applying an equation that determines energy as:

$Q = Δ m c^{2}$

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

Where the mass defect $Δ m$ is given by: $Δ m = m_{L_{3}^{7}} + m_{H_{1}^{1}} - m_{B e_{4}^{7}} - m_{n_{0}^{2}} c^{2}$ .

Put known values for finding Q

For calculating the Q for the reaction:

$\begin{array}{rcl} Q & = & (6.941 u + 1.0079 u - 7.01692 u - 1.00866 u) c^{2} \\ = & - 0.00176 \times 931.5 M e V \\ = & - 1.64 M e V \end{array}$

Therefore, the amount of energy released is $Q = - 1.64 M e V$

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Determine Q for the reaction ${}_{3}^{7}L i + {}_{1}^{1}H \to {}_{4}^{7}B e + {}_{0}^{1}n$

Short Answer

Step by step solution

Given data and concept

Put known values for finding Q

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Most popular questions from this chapter

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Determine Q for the reaction ${}_{3}^{7}L i + {}_{1}^{1}H \to {}_{4}^{7}B e + {}_{0}^{1}n$