# A stationary uranium nucleus (238U) decay with emission of alpha particle and total kinetic energy. 238/92U ↣234/90Th+ 4/2a what is Kinetic energy of patricle?

Nov 28, 2017

${K}_{\setminus \alpha} = 5.203$ $M e V = 8.336 \setminus \times {10}^{- 13}$ $J$

#### Explanation:

${U}_{92}^{238} \setminus \rightarrow T {h}_{90}^{234} + \setminus {\alpha}_{2}^{4}$

Momentum Conservation: $\setminus \quad {\vec{P}}_{i} = {\vec{P}}_{f}$
${\vec{P}}_{U} = {\vec{P}}_{T h} + {\vec{P}}_{\setminus \alpha}$

The Uranium nucleus is at rest (${\vec{P}}_{U} = \vec{0}$).
vec P_{Th} + vec P_{\alpha} = vec 0; \qquad vec P_{\alpha} = - vec P_{Th}; \qquad P_{\alpha} = P_{Th} = P_0

Kinetic Energies:
K_{\alpha} = P_{\alpha}^2/(2m_{\alpha}); \qquad K_{Th} = P_{Th}^2/(2m_{Th}^234);

P_{\alpha} = P_{Th} = P_0;

m_{U}^238 = 238.050788 \quad u; \qquad m_{Th}^234 = 234.04360123 \quad u;
m_{\alpha} = 4.001506 \quad u;

${K}_{\setminus \alpha} / {K}_{T h} = {\left({P}_{\setminus \alpha} / {P}_{T h}\right)}^{2} \left({m}_{T h} / {m}_{\setminus \alpha}\right) = {m}_{T h} / {m}_{\setminus \alpha} = \frac{234.0436 \setminus \quad u}{4.001 \setminus \quad u} = 58.49$

K_{Th} = K_{\alpha}/58.49 = 0.017095.K_{\alpha}; ...... (1)

Fission Energy - Q Value:

$Q = \setminus \Delta m . {c}^{2} = \left[{m}_{U} - \left({m}_{T h} + {m}_{\setminus \alpha}\right)\right] . {c}^{2} = 0.005681 u . {c}^{2} =$

To convert the atmoc mass unit (u) to $M e V$ use the conversion factor,
$1 u = 931.5$ $\frac{M e V}{c} ^ 2$

Thus the energy released in the fission is, $\setminus \quad Q = 5.292$ $M e V$. This energy is distributed between the alpha particle and the Thorium-234 nucleus as kinetic energies.

${K}_{\setminus \alpha} + {K}_{T h} = Q = 5.292$ $M e V$ ...... (2)

Substituting (1) in (2) to eliminate ${K}_{T h}$,

$\left(1 + 0.017095\right) {K}_{\setminus \alpha} = 5.292$ $M e V$

${K}_{\setminus \alpha} = 5.203$ $M e V = 8.336 \setminus \times {10}^{- 13}$ $J$