Question

50 points QUIZLET ( i dont have an account, but there is answer to this question, if u dont want to bother solving it) Assume uranium-236 splits into two nuclei of palladium-117 (Pd). (The atomic mass of uranium is 236.0455561 u; that of palladium is 116.9178 u.) a. Write down the reaction. b. What other particles must be produced? c. What is the energy released?​

Answer 1

The conservation laws we can find the complete reaction of Uranium-235 and the correct answer is:

               $^{235}_{92}U + ^1_0n$  → $^{236}_{92} U*$

               $^{236}_{92}U*$  →  $2 \ ^{116}_{46} \ Pd + 2 \ ^1_0n + 195 \ MeV$

Nuclear fission occurs when a heavy nucleus is divided into two smaller nuclei, for the process to occur requires the absorption of some particle or some form of energy that creates an unstable heavy nucleus this nucleus passes to a lower energy configuration when be divided.

In the case of Uranium 235 this occurs when a neutron is absorbed, according to the reaction

                    $^{235}_{92} U + ^1_0n$ →  $^{236}_{92} U*$

Where uranium 235 and 236 of atomic mass, * indicates that the nucleus is unstable and decays after about 10⁻¹² s

 

Indicate that one of the daughter nuclei is palladium, which is why a partial reaction remains

                  ^{236}_{92} U* → $^{116}_{46}Pd$ + Y + $^1_0n$

 

Where Y is the other daughter nucleus and n the neutrons released in the reaction

The conservation laws must be fulfilled, let's use:  the conservation of atomic number and mass

.

1) The conservation of the atomic number allows to find the number of protons (atomic number) to atoms Y

                #_protons = 92- 46

                #_protons = 46 protons

The other atom is also a Palladium atom, the reaction is

           ^{236}_{92} U* → $2 \ ^{116}_{46}Pd$  + $^1_0n$

2) Conservation of mass we can find the number of neutrons emitted in the reaction

The mass of the palladium atoms is

                   m_ {children} = 2  116.9178 u

                   m_ {children} = 233.8256 u

The mass difference between the uranium and the daughter nuclei is

                   Δm = m_ u - m_ {daughters}

                   Δm = 236.0455561 - 233.8256

                   Δm = 2.2199561 u

The mass of a neutron tabulated is m_n = 1.000866492 u

Consequently there is only mass for two neutrons that must be emitted in the process, the mass of the two neutrons is

                  m_ {total neutron} = 2.001732984 u

The amount of mass remaining is converted into energy

                   ΔE = m c²

                           

The remaining mass to be convert to energy is

                 m_{remaining} = 2.2199561 - 2.001732984 = 0.2103626 u

                  m = 0.2102626 u (1.66054 10⁻²⁷ kg / 1u) = 0.3491495 10⁻²⁷ kg

                  ΔE = 0.3491495 10⁻²⁷ (2.99 10⁸) ²

                  ΔE = 3.1214 10⁻¹¹ J

It is more common to give this energy in eV

                 ΔE = 3.1214 10⁻¹¹ J ( $\frac{1 eV}{1.6 \ 10^{-19} J}$ ) = 1.9509 10⁸ eV

We summarize the complete reaction is

               $^{235}_{92}U + ^1_0n$  → $^{236}_{92} U*$

               $^{236}_{92}U*$  →  $2 \ ^{116}_{46} \ Pd + 2 \ ^1_0n + 195 \ MeV$

We can see that this reaction is not very efficient since a neutron is needed to start it and only two neutrons are emitted at the end of the process, even when the amount of energy emitted is high 195 MeV

In conclusion Using the conservation laws we can shorten the complete reaction of Uranium giving that the correct answer is:

                  $^{235}_{92}U + ^1_0n$  → $^{236}_{92} U*$

               $^{236}_{92}U*$  →  $2 \ ^{116}_{46} \ Pd + 2 \ ^1_0n + 195 \ MeV$

Learn more about nuclear fission here:

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