Question: Consider the following nuclear disintegration. \( {}_{\text{Z}}^{\text{A}}X \to {}_{{\text{Z - 2}...

Question

Consider the following nuclear disintegration.
${}_{\text{Z}}^{\text{A}}X \to {}_{{\text{Z - 2}}}^{{\text{A + k}}}X + {}_2^4He$ . Then the value of $k$ will be:
(A) $4$
(B) $2$
(C) $- 4$
(D) $- 2$

Answer 1

Solution

Hint : To solve this question, we need to use the conservation of the mass and the charge on both sides of the equation given. The mass is indicated by the mass numbers of the elements present, and the charge is indicated by the atomic numbers.

Complete step by step answer
In a nucleus, the mass is contributed by the protons and the neutrons, which make up the nucleus. The neutrons and the protons are collectively known as the nucleons. The mass of the neutron is nearly equal to that of the proton. So the total number of nucleons represents the mass number of a nucleus. The mass number or the total number of nucleons is indicated by the number present on the top side of the symbol of the element to which the nucleus belongs.
Also, the charge of a nucleus is contributed by the protons present. We know that the protons carry positive charge. So the total number of protons represents the atomic number of the nucleus.
We know that in any chemical reaction, the mass and the charge is conserved. So we can conserve the mass number and the atomic number in the nuclear disintegration which is given by the equation
$\Rightarrow {}_{\text{Z}}^{\text{A}}X \to {}_{{\text{Z - 2}}}^{{\text{A + k}}}X + {}_2^4He$
The total mass number on the left hand side is $A$ . Also the total mass number on the right hand side is equal to $A + k + 4$ . Conserving the mass number on both sides of the equation we get
$\Rightarrow A = A + k + 4$
Subtracting $A$ from both the sides we get
$\Rightarrow k + 4 = 0$
$\Rightarrow k = - 4$
Thus the value of $k$ is $- 4$ .

Note
We might argue that in the nuclear disintegration there is a release of the binding energy, which causes the total mass of the nuclei on the right hand side of the equation to be less than that on the left hand side. This difference in masses is called mass defect. But there is no information about the energy release given in the question. So we have neglected the mass defect in this question.