Question

Question: 13. Complete each of the following equations by adding the missing species:

\(\begin{array}{l}{\bf{(a) }}_{{\bf{13}}}^{{\bf{27}}}{\bf{Al + }}_{\bf{2}}^{\bf{4}}{\bf{Hen }} \to {\bf{? + }}_{\bf{0}}^{\bf{1}}{\bf{n}}\\{\bf{(b) }}_{{\bf{94}}}^{{\bf{239}}}{\bf{Pu + }} \to {\bf{?n}}_{{\bf{96}}}^{{\bf{242}}}{\bf{Cm + }}_{\bf{0}}^{\bf{1}}{\bf{n}}\\{\bf{(c) }}_{{\bf{ 7}}}^{{\bf{14}}}{\bf{\;N + }}_{\bf{2}}^{\bf{4}}{\bf{Hen }} \to {\bf{? + }}_{\bf{1}}^{\bf{1}}{\bf{H}}\\{\bf{(d) }}_{{\bf{92}}}^{{\bf{235}}}{\bf{Un }} \to {\bf{? + }}_{{\bf{55}}}^{{\bf{135}}}{\bf{Cs + 4}}_{\bf{0}}^{\bf{1}}{\bf{n}}\end{array}\)

Short answer

\(\begin{array}{l}a) _{13}^{27}Al + _2^4He \to _{15}^{30}P + _0^1n\\b) _{94}^{239}Pu + _2^4He \to _{96}^{242}Cm + _0^1n\\c) _7^{14}\;N + _2^4He \to _8^{17}O + \frac{1}{1}H\\d) _{92}^{235}U \to _{37}^{96}Rb + _{55}^{135}Cs + 4\frac{1}{0}n\end{array}\)

Step-by-step solution

Subpart a)

We have:\(_{13}^{27}Al + _2^4He \to _y^x? + _0^1n.\)

The sum of the masses on the left side is:

\(27 + 4 = 31\)

Which means on the right side we also need to have the sum of 31 :

\(\begin{array}{l}x + 1 = 31\\x = 30\end{array}\)

The sum of the atomic number on the left is:

13+2=15

Which means on the right side we also need to have the sum of 15:

\(\begin{array}{l}y + 0 = 15\\y = 15\end{array}\)

From the periodic table we can see that element which has an atomic number of 15 is P, which means the complete reaction is:

\(_{13}^{27}Al + _2^4He \to _{15}^{30}P + _0^1n\)

Subpart b)

We have:\(_{94}^{239}Pu + _y^x? \to _{96}^{242}Cm + _0^1n\)

The sum of the masses on the right side is:

242+1=243

Which means on the left side we also need to have the sum of 243 :

\(\begin{array}{l}239 + x = 243\\ x = 4\end{array}\)

The sum of the atomic number on the right is:

96+0=96

Which means on the left side we also need to have the sum of 96:

94+y=96

y=2

From the periodic table we can see that element which has an atomic number of 2 is He, which means the complete reaction is:

\(_{94}^{239}Pu + _2^4He \to _{96}^{242}Cm + _0^1n\)

Subpart c)

We have:\(_7^{14}N + _2^4He \to _y^x? + \frac{1}{1}H\)

The sum of the masses on the left side is:

14+4=18

Which means on the right side we also need to have the sum of 18 :

x+1=18

x=17

The sum of the atomic number on the left is:

7+2=9

Which means on the right side we also need to have the sum of 9:

y+1=9

y=8

From the periodic table we can see that element which has an atomic number of 8 is\(O,\)which means the complete reaction is:

\(_7^{14}\;N + _2^4He \to _8^{17}O + _1^1H\)

Subpart d)

We have:\(_{92}^{235}U \to _y^x? + _{55}^{135}Cs + 4\frac{1}{0}n\)

The mass on the left side is 235 , which means on the right side we also need to have the sum of 235 :

\(\begin{array}{l}x + 135 + 4 \times 1 = 235\\x = 96\end{array}\)

The atomic number on the left is 92 , which means on the right side we also need to have the sum of 92 :

\(\begin{array}{l}y + 55 + 4 \times 0 = 92\\y = 37\end{array}\)

From the periodic table we can see that element which has an atomic number of 37 is\(Rb,\)which means the complete reaction is:

\(_{92}^{235}U \to _{37}^{96}Rb + _{55}^{135}Cs + 4_0^1n\)