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Nuclear Fusion

A Find the Q ...

Question

A) Find the Q - value and the kinetic energy of the emitted $α$ -particle in the $α$ -decay of $\begin{matrix} 22688 \end{matrix} R a$ .

B) Find the Q - value and the kinetic energy of the emitted $α$ -particle in the $α$ -decay of $\begin{matrix} 22688 \end{matrix} R a$ .
Given,
$m (\begin{matrix} 22688 \end{matrix} R a) = 226.02540 u,$
$m (\begin{matrix} 22286 \end{matrix} R n) = 222.01750 u,$
$m (\begin{matrix} 22086 \end{matrix} R n) = 220.01137 u,$
$m (\begin{matrix} 21684 \end{matrix} P o) = 216.00189 u,$

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Solution

A) $α$ - particle decay of $\begin{matrix} 22688 \end{matrix} R a$ emits a helium nucleus. As a result, its mass number reduces to 222 ( 226 - 4) and its atomic number reduces to 86 (88-2). This is shown in the following nuclear reaction
$\begin{matrix} 22688 \end{matrix} R a \to \begin{matrix} 22286 \end{matrix} R n + \begin{matrix} 42 \end{matrix} H e$
Q - value of emitted $α$ - particle = (Sum of initial mass - Sum of final mass) $c^{2}$
Where, c = Speed of light
It is given that : $m (226 R a) = 226.02540 u$
$m (\begin{matrix} 22686 \end{matrix} R n) = 222.01750$
Q - value
$= [226.02540 - (222.01750 + 4.002603)] u c^{2}$
$= 0.005297 u c^{2}$
$= 0.005297 \times 931.51 \approx 4.93 MeV$
Kinetic energy of the a - particle
$= (\frac{Mass number after decay}{Mass number before decay}) \times Q$
$= \frac{222}{226} \times 4.93$
$= 4.85 MeV$
Final Answer : $Q = 4.93 MeV$ ,
$E_{α} = 4.85 MeV$

B) $α$ - particle decay of $\begin{matrix} 22086 \end{matrix} R n$
$\begin{matrix} 22086 \end{matrix} \to \begin{matrix} 21684 \end{matrix} P o + \begin{matrix} 42 \end{matrix} H e$
it is given that:
Mass of $\begin{matrix} 22086 \end{matrix} R n = 220.01137 u$
Mass of $\begin{matrix} 21684 \end{matrix} P o = 216.00189 u$
Q - value
$= 220.01137 - (216.00189 + 4.002603)] u c^{2}$
$= 0.00688 u c^{2}$
$Q = 0.00688 \times 931.5 = 6.41 MeV$
Kinetic energy of the $α$ particle = $= \frac{220 - 4}{220} \times 6.41$ = 6.29 MeV
Final Answer : Q = 6.41 MeV
$E_{α} = 6.29 MeV$

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Similar questions

Find the Q-value and the kinetic energy of the emitted

α

- Particle in the

α

_{86}^{220} R n

_{88}^{226} R a

m (_{88}^{226} R a) = 226.02540 u

,

m (_{86}^{222} R n) = 222.01750 u

,

m (_{84}^{216} P o) = 216.00189 u

Q. Find the Q-value and the kinetic energy of the emitted α-particle inthe α-decay of (a) 22688 Ra and (b) 22086 Rn .Given m ( 22688 Ra ) = 226.02540 u, m ( 22286 Rn ) = 222.01750 u,m ( 22286 Rn ) = 220.01137 u, m ( 21684 Po ) = 216.00189 u.

Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) and (b).

Given = 226.02540 u, = 222.01750 u,

= 220.01137 u, = 216.00189 u.

Q. The kinetic energy of

α - p a r t i c l e

α - d e c a y

_{88} R a^{226}

is [given, mass number of Ra=222 u]

_{235}^{239} P u_{94}

α

- decay according to the equation

_{94}^{235} P u \to_{97}^{235} U +_{2}^{4} H e

. The energy released in the process is mostly kinetic energy of the

α

- particle. However, a part of the energy is released as

γ -

rays. What is the speed of the emitted

α -

particle if the

γ - r a y s

radiated out have energy of 0.90 MeV? Given: Mass of

_{94}^{239} P u = 239.05122 u

_{97}^{235} U = 235.04299 u

_{2}^{4} H e = 4.002602 u

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