A Plutonium-239 nucleus ({^{239}_{94}\text{Pu}}) captures a thermal neutron ({^{1}_{0}\tex… — JAMB UTME

A Plutonium-239 nucleus (

{^{239}_{94}\text{Pu}}

) captures a thermal neutron (

{^{1}_{0}\text{n}}

) and undergoes nuclear fission according to the reaction equation:

{^{239}_{94}\text{Pu}} + {^{1}_{0}\text{n}} \rightarrow {^{144}_{56}\text{Ba}} + {^{93}_{38}\text{Sr}} + x\,{^{1}_{0}\text{n}} + Q

Given the rest masses:
- Mass of ${^{239}_{94}\text{Pu}} = 239.0522\text{ u}$
- Mass of ${^{1}_{0}\text{n}} = 1.0087\text{ u}$
- Mass of ${^{144}_{56}\text{Ba}} = 143.9229\text{ u}$
- Mass of ${^{93}_{38}\text{Sr}} = 92.9154\text{ u}$
- $1\text{ u} = 931.5\text{ MeV}$

What is the number of emitted neutrons $x$ and the total energy $Q$ released in this fission process?

$x = 3$ neutrons and $Q = 183.04\text{ MeV}$ Answer
B
$x = 3$ neutrons and $Q = 756.56\text{ MeV}$
C
$x = 2$ neutrons and $Q = 183.04\text{ MeV}$
D
$x = 3$ neutrons and $Q = 2.11 \times 10^{-4}\text{ MeV}$

Answer

The reaction releases

x = 3

neutrons and an energy

Q = 183.04\text{ MeV}

The conservation of nucleon number requires

239 + 1 = 144 + 93 + x

, giving

x = 3

neutrons. The mass defect is calculated by subtracting the mass of products (

143.9229 + 92.9154 + 3 \times 1.0087 = 239.8644\text{ u}

) from the mass of reactants (

239.0522 + 1.0087 = 240.0609\text{ u}

), yielding

\Delta m = 0.1965\text{ u}

. Converting this to energy gives

Q = 0.1965 \times 931.5\text{ MeV} = 183.04\text{ MeV}

Step-by-Step Solution

Balance mass number (

A

) to find the number of neutrons

x

239 + 1 = 144 + 93 + x \implies 240 = 237 + x \implies x = 3

Total mass number must be conserved in a nuclear reaction.

Calculate the total mass of the reactants (

m_{\text{reactants}}

)

m_{\text{reactants}} = 239.0522\text{ u} + 1.0087\text{ u} = 240.0609\text{ u}

The reactants consist of one Plutonium-239 nucleus and one thermal neutron.

Calculate the total mass of the products (

m_{\text{products}}

)

m_{\text{products}} = 143.9229\text{ u} + 92.9154\text{ u} + 3(1.0087\text{ u}) = 239.8644\text{ u}

The products consist of one Barium-144 nucleus, one Strontium-93 nucleus, and three neutrons.

Calculate the mass defect (

\Delta m

)

\Delta m = 240.0609\text{ u} - 239.8644\text{ u} = 0.1965\text{ u}

Mass defect is the difference between total mass of reactants and total mass of products.

Calculate the liberated energy (

Q

)

Q = 0.1965\text{ u} \times 931.5\text{ MeV/u} = 183.03975\text{ MeV} \approx 183.04\text{ MeV}

Using Einstein's mass-energy equivalence conversion factor

1\text{ u} = 931.5\text{ MeV}

Key Concept

Nuclear fission conservation laws (mass/atomic number conservation) and mass defect energy equivalence (

Q = \Delta m \cdot 931.5\text{ MeV/u}

Estimated Time:2m 0s

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