Question: 14. Complete each of the following equations:

\(\begin{array}{l}{\bf{(a) }}_{\bf{3}}^{\bf{7}}{\bf{Li + ?}} \to {\bf{2}}_{\bf{2}}^{\bf{4}}{\bf{He}}\\\\{\bf{(b) }}_{\bf{6}}^{{\bf{14}}}{\bf{C}} \to _{\bf{7}}^{{\bf{14}}}{\bf{\;N + ?}}\\\\{\bf{(c) }}_{{\bf{13}}}^{{\bf{27}}}{\bf{Al + }}_{\bf{2}}^{\bf{4}}{\bf{He}} \to {\bf{? + }}_{\bf{0}}^{\bf{1}}{\bf{n}}\\\\{\bf{(d) }}_{{\bf{96}}}^{{\bf{250}}}{\bf{Cm}} \to {\bf{? + }}\,\,_{{\bf{38}}}^{{\bf{98}}}{\bf{Sr + 4}}_{\bf{0}}^{\bf{1}}{\bf{n}}\end{array}\)

We have:\(_3^7Li + _y^x? \to 2_2^4He\)

The mass on the right side is:

\(4.4 = 8\)

Which means on the left side we also need to have the sum of 8 :

\(\begin{array}{l}7 + x = 8\\x = 1\end{array}\)

The atomic number on the right is:

\(2 \times 2 = 4\)

Which means on the left side we also need to have the sum of 4 :

\(\begin{array}{l}3 + y = 4\\y = 1\end{array}\)

From the periodic table we can see that element which has an atomic number of 1 is\(H,\) which means the complete reaction is:

\(_3^7Li + _1^1H \to _2^4He\)

We have:\(_6^{14}C \to _7^{14}N + _y^x?\)

The mass on the left side is 14 which means on the right side we also need to have the sum of 14:

\(x + 14 = 14\)

x=0

The atomic number on the left is 6 , which means on the left side we also need to have the sum of 6 :

\(\begin{array}{l}7 + y = 6\\y = - 1\end{array}\)

We know that element which has an atomic number of -1 and mass number 0 is electron, which means the complete reaction is:

\(_6^{14}C \to _7^{14}N + _{ - 1}^0e\)

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

The sum of the masses on the right side is:

27+4=31

Which means on the left 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 right is:

13+2=15

Which means on the left 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:

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

We have:\(_{96}^{250}Cm \to _y^x? + _{38}^{98}Sr + 4_0^1n\)

The mass on the left side is 250 , which means on the right side we also need to have the sum of 250 :

\(\begin{array}{l}x + 98 + 4 \times 1 = 250\\x = 148\end{array}\)

The atomic number on the left is 96 , which means on the left side we also need to have the sum of 96 :

\(\begin{array}{l}y + 38 + 4 \times 0 = 96\\y = 58\end{array}\)

From the periodic table we can see that element which has an atomic number of 58 is Ce, which means the complete reaction is:

\(_{96}^{250}Cm \to _{58}^{148}Ce + _{38}^{98}Sr + 4_0^1n\)