Question

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.

Short answer

• a)The identity of X for this reaction is boron ${}^{9}B$.
• b)The identity of X for this reaction is nitrogen ${}^{13}N$.
• c) The identity of X for this reaction is carbonrole="math" localid="1663084579220" ${}^{12}C$.

Step-by-step solution

Step 1: Stating given data

The given reactionsare

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

Understanding concept of nuclear reactions and charge conservation

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.

a) Solution for finding element X

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 ${}^{9}B$.

b) Solution for finding element X

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 ${}^{13}N$.

c) Solution for finding element X

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 $C^{12}$.