Chapter 44, Problem 35P

To determine

Which decay would occur.

Answer to Problem 35P

Option (c) $\text{N}\text{d}\to \text{H}\text{e}+\text{G}\text{e}$

Explanation of Solution

Case a:

Write the expression to calculate the Q value for the $e^{+}$ decay.

  $Q=m\left(\text{C}\text{a}\right)-2m\left(e^{+}\right)-m\left(\text{K}\right)$

Here, Q is the Q value for the reaction, $m\left(\text{C}\text{a}\right)$ is the mass of $\text{C}\text{a}$, $m\left(e^{+}\right)$ is the mass of the $e^{+}$ and $m\left(\text{K}\right)$ is the mass of $\text{K}$.

Substitute $39.962591\text{u}$ for $m\left(\text{C}\text{a}\right)$, $39.963999\text{u}$ for $m\left(\text{K}\right)$ and $0.000549\text{u}$ for $m\left(e^{+}\right)$ in the above equation to calculate Q.

  $\begin{array}{c}Q=39.962591\text{u}-39.963999\text{u}-2\left(0.000549\text{u}\right)\\ =-0.002497\text{u}\left(\frac{931.5\text{MeV}}{1\text{u}}\right)\\ =-2.33\text{MeV}\end{array}$

Since this reaction has negative Q value, this reaction never occurs.

Case b:

Write the expression to calculate the Q value for the decay.

  $Q=m\left(\text{R}\text{u}\right)-m\left(\text{H}\text{e}\right)-m\left(\text{M}\text{o}\right)$

Here, Q is the Q value for the reaction, $m\left(\text{R}\text{u}\right)$ is the mass of $\text{R}\text{u}$, $m\left(\text{H}\text{e}\right)$ is the mass of the $\text{H}\text{e}$ and $m\left(\text{M}\text{o}\right)$ is the mass of $\text{M}\text{o}$.

Substitute $97.905287\text{u}$ for $m\left(\text{R}\text{u}\right)$, $4.002603\text{u}$ for $m\left(\text{H}\text{e}\right)$ and $93.905088\text{u}$ for $m\left(\text{M}\text{o}\right)$ in the above equation to calculate Q.

  $\begin{array}{c}Q=97.905287\text{u}-4.002603\text{u}-93.905088\text{u}\\ =-0.002404\text{u}\left(\frac{931.5\text{MeV}}{1\text{u}}\right)\\ =-2.24\text{MeV}\end{array}$

Since this reaction has negative Q value, this reaction never occurs.

Case c:

Write the expression to calculate the Q value for the decay.

  $Q=m\left(\text{N}\text{d}\right)-m\left(\text{H}\text{e}\right)-m\left(\text{G}\text{e}\right)$

Here, Q is the Q value for the reaction, $m\left(\text{N}\text{d}\right)$ is the mass of $\text{N}\text{d}$, $m\left(\text{H}\text{e}\right)$ is the mass of the $\text{H}\text{e}$ and $m\left(\text{G}\text{e}\right)$ is the mass of $\text{G}\text{e}$.

Substitute $143.910083\text{u}$ for $m\left(\text{N}\text{d}\right)$, $4.002603\text{u}$ for $m\left(\text{H}\text{e}\right)$ and $139.905434\text{u}$ for $m\left(\text{G}\text{e}\right)$ in the above equation to calculate Q.

  $\begin{array}{c}Q=143.910083\text{u}-4.002603\text{u}-139.905434\text{u}\\ =0.002046\text{u}\left(\frac{931.5\text{MeV}}{1\text{u}}\right)\\ =1.91\text{MeV}\end{array}$

Since this reaction has positive Q value, this reaction will occurs.

Conclusion:

Since the reaction $\text{N}\text{d}\to \text{H}\text{e}+\text{G}\text{e}$ has positive Q value these would occur. Thus, option (c) is correct.

In the case of $e^{+}$, the Q value is negative and this reaction never occurs. Thus, option (a) is incorrect.

In the case of alpha decay, the Q value is negative and this reaction never occurs. Thus, option (b) is incorrect.