Chapter 40, Problem 42P

(a)

To determine

The Q value for the given reaction.

Explanation of Solution

Given:

The reaction:

  $\text{H}\text{+}\text{H}\to \text{H}\text{+}\text{H}\text{+}Q$

Formula used:

Write the reaction.

  $\text{H}\text{+}\text{H}\to \text{H}\text{+}\text{H}\text{+}Q$

Write the expression for the $Q$ value, in MeV, for the decay.

  $Q=\left(m_i-m_f\right)c^2\left(\frac{\text{931}{\text{.5MeVc}}^{\text{-2}}}{\text{u}}\right)$ .......... (1)

Here, $m_i$ is the mass of reactants and $m_f$ is the mass of products.

Write the expression for the mass of reactants.

  $m_i=m_{\text{H}}+m_{\text{H}}$ .......... (2)

Here, $m_{\text{H}}$ is the mass of $\text{H}$ .

Write the expression for the mass of products.

  $m_f=m_{\text{H}}\text{+}{m}_{{}^1\text{H}}$ .......... (3)

Here, $m_{\text{H}}$ is the mass of $\text{H}$ and $m_{{}^1\text{H}}$ is the mass of $\text{H}$ .

Calculation:

Substitute $\text{2}\text{.014102u }$ for $m_{\text{H}}$ equation (2).

  $\begin{array}{c}{m}_i=\text{2}\text{.014102u }+\text{2}\text{.014102u }\\ =4.028204\text{u}\end{array}$

Substitute $3.016049\text{u}$ for $m_{\text{H}}$ and $1.007825\text{u}$ for $m_{\text{n}}$ in equation (3).

  $\begin{array}{c}{m}_f=3.016049\text{u+}1.007825\text{u}\\ =4.028204\text{u}\end{array}$

Substitute $4.028204\text{u}$ for $m_i$ and $4.028204\text{u}$ for $m_f$ in equation (1).

  $\begin{array}{c}Q=\left(4.028204\text{u}-4.028204\text{u}\right)c^2\left(\frac{\text{931}{\text{.5MeVc}}^{\text{-2}}}{\text{u}}\right)\\ =\left(\text{0}\text{.004330u}\right)c^2\left(\frac{\text{931}{\text{.5MeVc}}^{\text{-2}}}{\text{u}}\right)\\ =4.03\text{MeV}\end{array}$

Conclusion:

Thus, the Q value for the given reaction is $4.03\text{MeV}$ .

(b)

To determine

The Q value for the given reaction.

Explanation of Solution

Given:

The reaction:

  $\text{H}\text{+}\text{H}\text{e}\to \text{H}\text{e+}\text{H}\text{+}Q$

Formula used:

Write the reaction.

  $\text{H}\text{+}\text{H}\text{e}\to \text{H}\text{e+}\text{H}\text{+}Q$

Write the expression for the $Q$ value, in MeV, for the decay.

  $Q=\left(m_i-m_f\right)c^2\left(\frac{\text{931}{\text{.5MeVc}}^{\text{-2}}}{\text{u}}\right)$ .......... (1)

Here, $m_i$ is the mass of reactants and $m_f$ is the mass of products.

Write the expression for the mass of reactants.

  $m_i=m_{\text{H}}+m_{\text{H}\text{e}}$ .......... (2)

Here, $m_{\text{H}}$ is the mass of $\text{H}$ and $m_{\text{H}\text{e}}$ is the mass of $\text{H}\text{e}$ .

Write the expression for the mass of products.

  $m_f=m_{\text{H}\text{e}}\text{+}{m}_{{}^1\text{H}}$ .......... (3)

Here, $m_{\text{H}\text{e}}$ is the mass of $\text{H}\text{e}$ and $m_{{}_{{}^1\text{H}}}$ is the mass of ${}^1\text{H}$ .

Calculation:

Substitute $\text{2}\text{.014102u }$ for $m_{\text{H}}$ and $3.016029\text{u}$ for $m_{\text{H}\text{e}}$ in equation (2).

  $\begin{array}{c}{m}_i=\text{2}\text{.014102u }+3.016029\text{u }\\ =\text{5}\text{.030131u}\end{array}$

Substitute $\text{4}\text{.002602u}$ for $m_{\text{H}\text{e}}$ and $1.007825\text{u}$ for $m_{{}_{{}^1\text{H}}}$ in equation (3).

  $\begin{array}{c}{m}_f=\text{4}\text{.002602u+}1.007825\text{u}\\ =\text{5}\text{.010427u}\end{array}$

Substitute $\text{5}\text{.030131u}$ for $m_i$ and $\text{5}\text{.010427u}$ for $m_f$ in equation (1).

  $\begin{array}{c}Q=\left(\text{5}\text{.030131u}-\text{5}\text{.010427u}\right)c^2\left(\frac{\text{931}{\text{.5MeVc}}^{\text{-2}}}{\text{u}}\right)\\ =\left(0.019703\text{u}\right)c^2\left(\frac{\text{931}{\text{.5MeVc}}^{\text{-2}}}{\text{u}}\right)\\ =18.4\text{MeV}\end{array}$

Conclusion:

Thus, the Q value for the given reaction is $18.4\text{MeV}$ .

(c)

To determine

The Q value for the given reaction.

Explanation of Solution

Given:

The reaction:

  $\text{L}\text{i+n}\to \text{H}\text{+}\text{H}\text{e+}Q$

Formula used:

Write the reaction.

  $\text{L}\text{i+n}\to \text{H}\text{+}\text{H}\text{e+}Q$

Write the expression for the $Q$ value, in MeV, for the decay.

  $Q=\left(m_i-m_f\right)c^2\left(\frac{\text{931}{\text{.5MeVc}}^{\text{-2}}}{\text{u}}\right)$ .......... (1)

Here, $m_i$ is the mass of reactants and $m_f$ is the mass of products.

Write the expression for the mass of reactants.

  $m_i=m_{\text{L}\text{i}}+m_{\text{n}}$ .......... (2)

Here, $m_{\text{L}\text{i}}$ is the mass of $\text{L}\text{i}$ and $m_{\text{n}}$ is the mass of $\text{n}$ .

Write the expression for the mass of products.

  $m_f=m_{\text{H}}\text{+}{m}_{\text{H}\text{e}}$ .......... (3)

Here, $m_{\text{H}}$ is the mass of $\text{H}$ and $m_{\text{H}\text{e}}$ is the mass of $\text{H}\text{e}$ .

Calculation:

Substitute $6.015122\text{u }$ for $m_{\text{L}\text{i}}$ and $\text{1}\text{.008665u}$ for $m_{\text{n}}$ in equation (2).

  $\begin{array}{c}{m}_i=6.015122\text{u +1}\text{.008665u}\\ =\text{7}\text{.023787u}\end{array}$

Substitute $\text{3}\text{.016049u}$ for $m_{\text{H}}$ and $\text{4}\text{.002602u}$ for $m_{\text{H}\text{e}}$ in equation (3).

  $\begin{array}{c}{m}_f=\text{3}\text{.016049u}+\text{4}\text{.002602u}\\ =\text{7}\text{.018651u}\end{array}$

Substitute $\text{7}\text{.023787u}$ for $m_i$ and $\text{7}\text{.018651u}$ for $m_f$ in equation (1).

  $\begin{array}{c}Q=\left(\text{7}\text{.023787u}-\text{7}\text{.018651u}\right)c^2\left(\frac{\text{931}{\text{.5MeVc}}^{\text{-2}}}{\text{u}}\right)\\ =\left(0.003510\text{u}\right)c^2\left(\frac{\text{931}{\text{.5MeVc}}^{\text{-2}}}{\text{u}}\right)\\ =3.27\text{MeV}\end{array}$

Conclusion:

Thus, the Q value for the given reaction is $3.27\text{MeV}$ .