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Chapter 21: Problem 6

Complete these nuclear equations and identify \(X\) in each case: (a) \({ }^{135}{ }_{53} \mathrm{I} \longrightarrow{ }_{54}^{135} \mathrm{Xe}+\mathrm{X}\) (b) \({ }_{19}^{40} \mathrm{~K} \longrightarrow{ }_{-1}^{0} \beta+\mathrm{X}\) (c) \({ }_{27}^{59} \mathrm{Co}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{25}^{56} \mathrm{Mn}+\mathrm{X}\) (d) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{40}^{99} \mathrm{Zr}+{ }_{52}^{135} \mathrm{Te}+2 \mathrm{X}\)

Short Answer

Expert verified
The identified unknown \(X\) in each case is: (a) Electron \(e^{-}\), (b) Calcium \(^{40}_{20}Ca\), (c) Alpha particle \(^{4}_{2}He\), (d) Neutron \(^{1}_{0}n\)

Step by step solution

01

Analyze the equations and fill missing values

For each equation, the aim should be to check that the numbers on both sides of the reaction equation add up. This means that the sum of atomic numbers (subscripts) and the sum of mass numbers (superscripts) should be the same on both sides of the equation.
02

Solve for X in first equation

For the equation \(^{135}_{53}I \rightarrow ^{135}_{54}Xe + X\), by comparing the superscripts, we notice that \(135=135 + mass~number~of~X\). This means that the mass number of X is 0. Comparing the subscripts, \(53 = 54 + atomic~number~of~X\), so the atomic number of X is -1. An atomic number of -1 and mass number of 0 corresponds to an electron. Hence, \(X\) is an electron.
03

Solve for X in second equation

For the equation \(^{40}_{19}K \rightarrow ^{0}_{-1}\beta + X\), we again similarly get mass number of \(X\) as \(40\) and atomic number as \(20\). A mass number of \(40\) and atomic number \(20\) corresponds to calcium (Ca). Hence, \(X\) is \(^{40}_{20}Ca\).
04

Solve for X in third equation

For \(^{59}_{27}Co + ^{1}_{0}n \rightarrow ^{56}_{25}Mn + X\), to solve for \(X\), we can see that the mass number of \(X\) will be \(59+1-56=4\), and atomic number will be \(27+0-25=2\). So, \(X\) will be an alpha particle \(^{4}_{2}He\).
05

Solve for X in fourth equation

The final equation is \(^{235}_{92}U + ^{1}_{0}n \rightarrow ^{99}_{40}Zr + ^{135}_{52}Te + 2X\). Here, the mass number of \(X\) will be \((235+1-99-135)/2=1\), and the atomic number will be \((92+0-40-52)/2=0\). Therefore, \(X\) will be a neutron \(^{1}_{0}n\).

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Most popular questions from this chapter

In the thorium decay series, thorium- 232 loses a total of \(6 \alpha\) particles and \(4 \beta\) particles in a 10 -stage process. What is the final isotope produced?

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