Chapter 30, Problem 36P

To determine

(a) To determine:

Recoil kinetic energy of daughter nucleus

Answer to Problem 36P

Solution:

$K_{Th}=0.0718MeV$

Explanation of Solution

Given Info:

${}_{92}{}^{238}U\to {}_{90}{}^{234}Th+{}_2{}^{4}He$

α particle emitted has 4.20 MeV kinetic energy.

Formula Used:

Kinetic Energy is given by

$K=\frac{p^2}{2m}$

Calculation:

The reaction from the given information is

${}_{92}{}^{238}U\to {}_{90}{}^{234}Th+{}_2{}^{4}He$

The momentum of the uranium nucleus is at rest initially.

Momentum of the Thorium is equal to momentum of Helium particle.

$\begin{array}{l}K=\frac{p^2}{2m}\\ K_{Th}=\frac{p_{Th}^2}{2m_{Th}}\\ p_{Th=}{p}_{\alpha }\\ K_{Th}=\frac{p_{{}_{\alpha }}^2}{2m_{Th}}\\ =\frac{{(m_{{}_{\alpha }}{v}_{{}_{\alpha }})}^2}{2m_{Th}}\\ =\frac{m_{{}_{\alpha }}}{m_{Th}}\frac{1}{2}{m}_{{}_{\alpha }}{v}_{{}_{\alpha }}^2\\ =\frac{m_{{}_{\alpha }}}{m_{Th}}{K}_{\alpha }\\ m_{{}_{\alpha }}=4u\\ m_{Th}=234u\end{array}$

Substituting the values

$K_{Th}=0.0718MeV$

To determine

(b) To determine:

Q value of the decay

Answer to Problem 36P

Solution:

$Q=4.2718MeV$

Explanation of Solution

Q value can be calculated as

$\begin{array}{l}Q=K_{\alpha }+K_{Th}\\ Q=(4.20+0.0718)MeV\\ Q=4.2718MeV\end{array}$