Question: Identify X in the following nuclear reactions:

(a) ${}^{1}H+{}^{9}Be\to X+n;\left(b\right){}^{12}C+{}^{1}H\to X;\left(c\right){}^{15}N+{}^{1}H\to {}^{4}He+X.$Appendix F will help.

The given reactionsare

• (a)${}^{1}H+{}^{9}Be\to X+n$
• (b)${}^{12}C+{}^{1}H\to X$
• (c)${}^{15}N+{}^{1}H\to {}^{4}He+X$.

In any nuclear reaction between two nuclei or a nucleus of a molecule reacting with an external substance, we get one or more nuclides with a certain amount of energy produced resulting in their reaction. This concept of nuclear reactions follows the important concept of charge conservation due to the laws of nature. Using this concept, the element can be found through its atomic number or charge.

Formula:

Mass number of an element, A = N+Z (i)

wherethenumber of protons (Z), the number of neutrons (N), and the number of electrons are each conserved.

For reaction (a):

${}^{1}H$has 1 proton, 1 electron, and 0 neutrons and${}^{9}Be$has 4 protons, 4 electrons, and 9 – 4 = 5 neutrons, so using equation (i), X has 1 + 4 = 5 protons, 1 + 4 = 5 electrons, and 0 + 5 – 1 = 4 neutrons. One of the neutrons is freed in the reaction. X must be boron with a molar mass of 5 + 4 = 9 g/mol. Hence, the element is boron ${}^{9}B$.

For reaction (b):

${}^{12}C$has 6 protons, 6 electrons, and 12 – 6 = 6 neutrons and${}^{1}H$has 1 proton, 1 electron, and 0 neutrons, so using equation (i), X has 6 + 1 = 7 protons, 6 + 1 = 7 electrons, and 6 + 0 = 6 neutrons.So, itmust be nitrogen with a molar mass of 7 + 6 = 13 g/mol.

Hence, the element is nitrogen ${}^{13}N$.

For reaction (c):

${}^{15}N$has 7 protons, 7 electrons, and 15 – 7 = 8 neutrons;${}^{1}H$has 1 proton, 1 electron, and 0 neutrons; and 4He has 2 protons, 2 electrons, and 4 – 2 = 2 neutrons; so, using equation (i) X has 7 + 1 – 2 = 6 protons, 6 electrons, and 8 + 0 – 2 = 6 neutrons.So, itmust be carbon with a molar mass of 6 + 6 = 12 g/mol.

Hence, the element is carbon $C^{12}$.