Chapter 13, Problem 45P

(a)

To determine

Check whether the decay ${}_{20}^{40}\text{Ca}\to {\text{e}}^{+}{+}_{19}^{40}\text{K+}v$ possible or not.

Answer to Problem 45P

Since the disintegration energy is negative, ${}_{20}^{40}\text{Ca}\to {\text{e}}^{+}{+}_{19}^{40}\text{K+}v$ cannot occur.

Explanation of Solution

Write the equation to find the disintegration energy.

    $Q=\left(m_{\text{initial}}-m_{\text{final}}\right)\left(931.5\text{MeV}/\text{u}\right)$

Here, $Q$ the disintegration energy, $m_{\text{initial}}$ the initial mass and $m_{\text{final}}$ the final mass.

Write the equation for $Q$ for ${}_{20}^{40}\text{Ca}\to {\text{e}}^{+}{+}_{19}^{40}\text{K+}v$.

    $Q=m\left({}_{20}^{40}\text{Ca}\right)-m\left({\text{e}}^{+}\right)-m\left({}_{19}^{40}\text{K}\right)$

Here, $m\left({}_{20}^{40}\text{Ca}\right)$ the atomic mass of ${}_{20}^{40}\text{Ca}$, $m\left({\text{e}}^{+}\right)$ the mass of positron and $m\left({}_{19}^{40}\text{K}\right)$ the mass of ${}_{19}^{40}\text{K}$.

Conclusion:

Substitute $39.96259\text{u}$ $m\left({}_{20}^{40}\text{Ca}\right)$, $0.0005486\text{u}$ for $m\left({\text{e}}^{+}\right)$, $39.96400\text{u}$ $m\left({}_{19}^{40}\text{K}\right)$ the above equation.

    $\begin{array}{c}Q=\left(39.96259\text{u}-0.0005486\text{u}-39.96400\text{u}\right)\left(931.5\text{MeV}/\text{u}\right)\\ =-1.82\text{MeV}\end{array}$

Negative value of $Q$ that the decay ${}_{20}^{40}\text{Ca}\to {\text{e}}^{+}{+}_{19}^{40}\text{K+}v$ not possible.

Thus, since the disintegration energy is negative, ${}_{20}^{40}\text{Ca}\to {\text{e}}^{+}{+}_{19}^{40}\text{K+}v$ cannot occur.

(b)

To determine

Check whether the decay ${}_{44}^{98}\text{Ru}{\to }_2^4\text{He}{+}_{42}^{94}\text{Mo}$ possible or not.

Answer to Problem 45P

Since the disintegration energy is negative, ${}_{44}^{98}\text{Ru}{\to }_2^4\text{He}{+}_{42}^{94}\text{Mo}$ cannot occur.

Explanation of Solution

Write the equation for $Q$ for ${}_{44}^{98}\text{Ru}{\to }_2^4\text{He}{+}_{42}^{94}\text{Mo}$.

    $Q=m\left({}_{44}^{98}\text{Ru}\right)-m\left({\text{e}}^{+}\right)-m\left({}_{19}^{40}\text{K}\right)$

Here, $m\left({}_{44}^{98}\text{Ru}\right)$ the atomic mass of ${}_{44}^{98}\text{Ru}$, $m\left({}_2^4\text{He}\right)$ the mass of ${}_2^4\text{He}$ and $m\left({}_{42}^{94}\text{Mo}\right)$ the mass of ${}_{42}^{94}\text{Mo}$.

Conclusion:

Substitute $97.9055\text{u}$ $m\left({}_{44}^{98}\text{Ru}\right)$, $4.0026\text{u}$ for $m\left({}_2^4\text{He}\right)$, $93.9047\text{u}$ $m\left({}_{42}^{94}\text{Mo}\right)$ the above equation.

    $\begin{array}{c}Q=\left(97.9055\text{u}-4.0026\text{u}-93.9047\text{u}\right)\left(931.5\text{MeV}/\text{u}\right)\\ =-1.68\text{MeV}\end{array}$

Negative value of $Q$ that the decay ${}_{44}^{98}\text{Ru}{\to }_2^4\text{He}{+}_{42}^{94}\text{Mo}$ not possible.

Thus, since the disintegration energy is negative, ${}_{44}^{98}\text{Ru}{\to }_2^4\text{He}{+}_{42}^{94}\text{Mo}$ cannot occur.

(c)

To determine

Check whether the decay ${}_{60}^{144}\text{Nd}{\to }_2^4\text{He}{+}_{58}^{140}\text{Ce}$ possible or not.

Answer to Problem 45P

Since the disintegration energy is possible, ${}_{60}^{144}\text{Nd}{\to }_2^4\text{He}{+}_{58}^{140}\text{Ce}$ can occur.

Explanation of Solution

Write the equation for $Q$ for Since the disintegration energy is possible, ${}_{60}^{144}\text{Nd}{\to }_2^4\text{He}{+}_{58}^{140}\text{Ce}$ can occur..

    $Q=m\left({}_{60}^{144}\text{Nd}\right)-m\left({}_2^4\text{He}\right)-m\left({}_{58}^{140}\text{Ce}\right)$

Here, $m\left({}_{60}^{144}\text{Nd}\right)$ the atomic mass of ${}_{60}^{144}\text{Nd}$, $m\left({}_2^4\text{He}\right)$ the mass of ${}_2^4\text{He}$ and $m\left({}_{58}^{140}\text{Ce}\right)$ the mass of $\left({}_{58}^{140}\text{Ce}\right)$.

Conclusion:

Substitute $143.9099\text{u}$ $m\left({}_{60}^{144}\text{Nd}\right)$, $4.0026\text{u}$ for $m\left({}_2^4\text{He}\right)$, $139.9054\text{u}$ $\left({}_{58}^{140}\text{Ce}\right)$ the above equation.

    $\begin{array}{c}Q=\left(143.9099\text{u}-4.0026\text{u}-139.9054\text{u}\right)\left(931.5\text{MeV}/\text{u}\right)\\ =1.86\text{MeV}\end{array}$

Positive value of $Q$ that the decay ${}_{60}^{144}\text{Nd}{\to }_2^4\text{He}{+}_{58}^{140}\text{Ce}$ possible.

Thus, since the disintegration energy is possible, ${}_{60}^{144}\text{Nd}{\to }_2^4\text{He}{+}_{58}^{140}\text{Ce}$ can occur.