Question

the nuclear physics test (2026)

1. (a) protons can be produced by the bombardment of nitrogen - 14 nuclei with alpha particles.

the nuclear reaction equation for this process is given below.
\\(_{7}^{14}\text{n} + _{2}^{4}\text{he} \to \text{x} + _{1}^{1}\text{h}\\)
identify the proton number and nucleon number for the nucleus x.
proton number:
nucleon number:
(b) the following data are available for the reaction in (a)
rest mass of nitrogen - 14 nucleus = 14.0031 u
rest mass of alpha particle = 4.0026 u
rest mass of x nucleus = 16.9991 u
rest mass of proton = 1.0073 u
show that the minimum kinetic energy that the alpha particle must have in order for the reaction to take place is about 0.7 mev.

Explanation:

Step1: Balance proton numbers

Proton number balance: $7 + 2 = Z + 1$, where $Z$ is proton number of X.
Solve for $Z$: $Z = 7+2-1=8$

Step2: Balance nucleon numbers

Nucleon number balance: $14 + 4 = A + 1$, where $A$ is nucleon number of X.
Solve for $A$: $A =14+4-1=17$

Step3: Calculate mass defect (reactants - products)

First, find total rest mass of reactants:
$m_{\text{reactants}} = 14.0031\ \text{u} + 4.0026\ \text{u} = 18.0057\ \text{u}$
Total rest mass of products:
$m_{\text{products}} = 16.9991\ \text{u} + 1.0073\ \text{u} = 18.0064\ \text{u}$
Mass defect $\Delta m = m_{\text{products}} - m_{\text{reactants}} = 18.0064 - 18.0057 = 0.0007\ \text{u}$

Step4: Convert mass defect to energy

Use $1\ \text{u} = 931.5\ \text{MeV}/c^2$, so kinetic energy needed:
$E = \Delta m \times 931.5\ \text{MeV} = 0.0007 \times 931.5 \approx 0.652\ \text{MeV} \approx 0.7\ \text{MeV}$

Answer:

(a)
Proton number: 8
Nucleon number: 17

(b)
The minimum kinetic energy of the alpha particle is approximately 0.7 MeV, as shown by the mass defect-energy conversion.