Complete and balance the nuclear equations for the following fission reactions: (a) ${ }_{99}^{239} \mathrm{Pu}+{ }_{0} \mathrm{n} \longrightarrow{ }_{52}^{137} \mathrm{Te}+{ }_{42}^{100} \mathrm{Mo}+$ (b) ${ }_{100}^{256} \mathrm{Fm}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{46}^{113} \mathrm{Pd}+{ }_{-}+4{ }_{0}^{1} \mathrm{n}$

Short Answer

Expert verified
The short answer is as follows: (a) \({ }_{99}^{239} \mathrm{Pu}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{52}^{137} \mathrm{Te}+{ }_{42}^{100} \mathrm{Mo}+{ }_{5}^3 \mathrm{B}\) (b) \({ }_{100}^{256} \mathrm{Fm}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{46}^{113} \mathrm{Pd}+{ }_{54}^{140} \mathrm{Xe}+4{ }_{0}^{1} \mathrm{n}\)

Step by step solution

01

Calculate atomic number imbalance

Find the difference between the total atomic numbers on both sides of the equation : Left side atomic numbers: 99 Right side atomic numbers: 52 + 42 = 94 Imbalance: 99 - 94 = 5
02

Calculate mass number imbalance

Find the difference between the total mass numbers on both sides of the equation : Left side mass numbers: 239 + 1 = 240 Right side mass numbers: 137 + 100 = 237 Imbalance: 240 - 237 = 3
03

Add the missing particle

We need to add a particle with atomic number 5 and mass number 3 to balance the equation. This particle is an isotope of boron, represented as \({ }_{5}^3 \mathrm{B}\): \({ }_{99}^{239} \mathrm{Pu}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{52}^{137} \mathrm{Te}+{ }_{42}^{100} \mathrm{Mo}+{ }_{5}^3 \mathrm{B}\) #b)# For the fission reaction: \({ }_{100}^{256} \mathrm{Fm}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{46}^{113} \mathrm{Pd}+{ }_{-}+4{ }_{0}^{1} \mathrm{n}\)
04

Calculate atomic number imbalance

Find the difference between the total atomic numbers on both sides of the equation : Left side atomic numbers: 100 Right side atomic numbers: 46 + (-) Imbalance: 100 - 46 = 54 (not counting the unknown element)
05

Calculate mass number imbalance

Find the difference between the total mass numbers on both sides of the equation : Left side mass numbers: 256 + 1 = 257 Right side mass numbers: 113 + 4 * 1 = 117 (not counting the unknown element) Imbalance: 257 - 117 = 140
06

Add the missing particle

We need to add a particle with atomic number 54 and mass number 140 to balance the equation. This particle is an isotope of xenon, represented as \({ }_{54}^{140} \mathrm{Xe}\): \({ }_{100}^{256} \mathrm{Fm}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{46}^{113} \mathrm{Pd}+{ }_{54}^{140} \mathrm{Xe}+4{ }_{0}^{1} \mathrm{n}\)

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Most popular questions from this chapter

Write balanced equations for (a) ${ }_{92}^{238} \mathrm{U}(\alpha, \mathrm{n}){ }^{241} \mathrm{Pu},$ (b) ${ }^{14} \mathrm{~N}(\alpha, \mathrm{p}){ }^{17} \mathrm{O},(\mathbf{c}){ }_{26}^{56} \mathrm{Fe}\left(\alpha, \beta^{-}\right)_{29}^{60} \mathrm{Cu} .$

Complete and balance the following nuclear equations by supplying the missing particle: (a) \({ }_{47}^{106} \mathrm{Ag}+{ }_{-1}^{0} \mathrm{e} \longrightarrow ?\) (b) \({ }_{106}^{263} \mathrm{Sg} \longrightarrow{ }_{2}^{4} \mathrm{He}+?\) (c) \({ }_{84}^{216} \mathrm{Po} \longrightarrow{ }_{82}^{212} \mathrm{~Pb}+?\) (d) ${ }_{5}^{10} \mathrm{~B}+? \longrightarrow{ }_{3} \mathrm{Li}+{ }_{2}^{4} \mathrm{He}$ (e) \({ }^{220} \mathrm{Rn} \longrightarrow{ }_{2}^{4} \mathrm{He}+?\)

A 10.00 -g plant fossil from an archaeological site is found to have a ${ }^{14} \mathrm{C}$ activity of 3094 disintegrations over a period of ten hours. A living plant is found to have a \({ }^{14} \mathrm{C}\) activity of 9207 disintegrations over the same period of time for an equivalent amount of sample with respect to the total contents of carbon. Given that the half-life of \({ }^{14} \mathrm{C}\) is 5715 years, how old is the plant fossil?

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Write balanced equations for each of the following nuclear reactions: $(\mathbf{a}){ }_{92}^{238} \mathrm{U}(\mathrm{n}, \gamma){ }^{239} \mathrm{U},(\mathbf{b}){ }_{82}^{16} \mathrm{O}(\mathrm{p}, \alpha){ }^{13} \mathrm{~N},\( (c) \){ }_{8}^{18} \mathrm{O}\left(\mathrm{n}, \beta^{-}\right){ }^{19} \mathrm{~F}$.

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