# Question 6abfe

Nov 4, 2016

$58.5$

#### Explanation:

We have the nuclear reaction
"_92^238$\text{U} \to$ "_90^234$\text{Th} +$"_2^4alpha

Let ${v}_{\alpha} \mathmr{and} {v}_{\text{Th}}$ be the velocities of the $\alpha$ particle and daughter nuclide respectively.

From conservation of momentum we get
$0 = {m}_{\text{Th" xxv_"Th}} + {m}_{\alpha} \times {v}_{\alpha}$
${m}_{\text{Th" xxv_"Th}} = - {m}_{\alpha} \times {v}_{\alpha}$
Notice that velocity of daughter nuclide is opposite to the velocity of $\alpha$ particle.

Squaring both sides
${\left({m}_{\text{Th" xxv_"Th}}\right)}^{2} = {\left(- {m}_{\alpha} \times {v}_{\alpha}\right)}^{2}$

Remembering that Kinetic energy of a body $K E = \frac{1}{2} m {v}^{2}$ and rearranging we get
 (m_alphav_alpha^2)/(m_"Th"v_"Th"^2)=m_"Th"/m_alpha

Multiply numerator and denominator of LHS by $\frac{1}{2}$
 (1/2m_alphav_alpha^2)/(1/2m_"Th"v_"Th"^2)=m_"Th"/m_alpha
 =>(KE_alpha)/(KE_"Th")=m_"Th"/m_alpha#
Inserting value of masses on the RHS we get
$\frac{K {E}_{\alpha}}{K {E}_{\text{Th}}} = \frac{234}{4} = 58.5$