A fission event studied at a research reactor sees a uranium-235 nucleus absorb a slow neutron and split according to the equation: ^10n+ ^23592U→ ^14054Xe+ ^94zSr+x ^10n where x additional neutrons are also released and z is the proton number of the strontium fragment. The nucl…

A Level Physics 9702: Balancing a Uranium-235 Fission Equation into Xenon and Strontium

A fission event studied at a research reactor sees a uranium-235 nucleus absorb a slow neutron and split according to the equation: $^1_0\text{n}+\,^{235}_{92}\text{U}\rightarrow\,^{140}_{54}\text{Xe}+\,^{94}_{z}\text{Sr}+x\,^1_0\text{n}$ where $x$ additional neutrons are also released and $z$ is the proton number of the strontium fragment.

The nuclear masses involved are $m(^{235}_{92}\text{U})=235.043930\text{ u}$ , $m(^1_0\text{n})=1.008665\text{ u}$ , $m(^{140}_{54}\text{Xe})=139.921646\text{ u}$ and $m(^{94}_{z}\text{Sr})=93.915356\text{ u}$ .

Take $1\text{ u} = 1.66\times10^{-27}\text{ kg} = 931.5\text{ MeV}$ and $c = 3.00\times10^8\text{ m s}^{-1}$ .

(a) By conserving nucleon number and proton number, find the value of $x$ and the value of $z$ . [2]

(b) Calculate the total mass of the particles before the reaction and the total mass of the particles after the reaction, and hence find the mass defect $\Delta m$ for this reaction, in u. [2]

(d) Calculate the energy released in this fission reaction, in joules. [2]

Part (a): Balancing the nuclear equation

The nucleon numbers (top numbers) on each side must be equal: $1+235=140+94+x\implies236=234+x\implies x=2$

The proton numbers (bottom numbers) on each side must also be equal (neutrons carry proton number 0): $0+92=54+z+2(0)\implies z=92-54=38$

Recompute as a check: with $x=2$ and $z=38$ , the nucleon numbers give $140+94+2(1)=236=1+235$ and the proton numbers give $54+38+2(0)=92=0+92$ , both sides balance exactly. (This also confirms the fragment is strontium, $Z=38$ , consistent with the symbol Sr already given.)

So $x=\boxed{2}$ neutrons are released, and $z=\boxed{38}$ .

Part (b): Mass defect

Total mass of the particles before the reaction (neutron $+$ uranium-235): $m_{before}=1.008665+235.043930=236.052595\text{ u}$

Total mass of the particles after the reaction (xenon-140 $+$ strontium-94 $+$ two neutrons): $m_{after}=139.921646+93.915356+2(1.008665)=139.921646+93.915356+2.017330=235.854332\text{ u}$

Mass defect: $\Delta m=m_{before}-m_{after}=236.052595-235.854332=0.198263\text{ u}$

Recompute as a check, combining differently: $(1.008665-2\times1.008665)+(235.043930-139.921646-93.915356)=(-1.008665)+(1.206928)=0.198263\text{ u}$ , the same result.

$\Delta m=\boxed{0.198}\text{ u (3 s.f.)}$

Part (c): Energy released, in MeV

Using the given conversion $1\text{ u}=931.5\text{ MeV}$ directly on the (unrounded) mass defect from (b): $E=\Delta m\times931.5=0.198263\times931.5$

Recompute as a check, splitting $0.198263$ as $0.2-0.001737$ : $0.2\times931.5-0.001737\times931.5=186.300-1.618=184.682\text{ MeV}$ , the same result as the direct multiplication.

$E=184.682\text{ MeV}$

$E=\boxed{185}\text{ MeV (3 s.f.)}$

Part (d): Energy released, in joules

Convert the mass defect to kilograms using $1\text{ u}=1.66\times10^{-27}\text{ kg}$ : $\Delta m=0.198263\times1.66\times10^{-27}=3.291\times10^{-28}\text{ kg}$

Apply $E=c^2\Delta m$ : $E=(3.00\times10^8)^2\times3.291\times10^{-28}=9.00\times10^{16}\times3.291\times10^{-28}$

$E=2.962\times10^{-11}\text{ J}$

Recompute as a check by converting the unrounded MeV answer from (c) directly using $1\text{ MeV}=1.602\times10^{-13}\text{ J}$ : $184.682\times1.602\times10^{-13}=2.959\times10^{-11}\text{ J}$ , consistent (small rounding only).

$E=\boxed{2.96\times10^{-11}}\text{ J (3 s.f.)}$

Final answers

(a) $x=\boxed{2}$ , $z=\boxed{38}$
(b) $\Delta m=\boxed{0.198}\text{ u}$
(c) $E=\boxed{185}\text{ MeV}$
(d) $E=\boxed{2.96\times10^{-11}}\text{ J}$