Question

A nuclear reaction of the kind \({ }_{2}^{3} \mathrm{He}+{ }_{6}^{12} \mathrm{C} \rightarrow \mathrm{X}+\alpha\) is called a pick-up nuclear reaction. a) Why is it called a pick-up reaction, that is, what is picked up, what picked it up, and where did it come from? b) What is the resulting nucleus X? c) What is the \(\mathrm{Q}\) -value of this reaction? d) Is this reaction endothermic or exothermic?

Short answer

Question: Explain the pick-up nuclear reaction involving helium and carbon nuclei, identify the resulting nucleus, determine the Q-value of the reaction, and identify if the reaction is endothermic or exothermic. Answer: In the given pick-up nuclear reaction, a helium nucleus (projectile) picks up nucleons from a carbon nucleus (target) to form a new nucleus. Here, the resulting nucleus (X) is identified as an isotope of carbon with a mass number of 11, represented as ${}_{6}^{11}\mathrm{C}$. The Q-value of the reaction is calculated to be approximately 2.88 MeV. Since the Q-value is positive, the reaction releases energy, making it an exothermic reaction.

Step-by-step solution

a) Explanation of a pick-up nuclear reaction

In a pick-up nuclear reaction, a projectile nucleus picks up a nucleon (or nucleons) from a target nucleus to form a new nucleus. In this case, the projectile nucleus is \({}_{2}^{3} \mathrm{He}\) (helium), the target nucleus is \({}_{6}^{12}\mathrm{C}\) (carbon), and the resulting nucleus X is the product of helium picking up a nucleon (or nucleons) from carbon.

b) Determination of the resulting nucleus X

Given the nuclear reaction: \({}_{2}^{3} \mathrm{He}+{ }_{6}^{12}\mathrm{C} \rightarrow \mathrm{X}+\alpha\) , we can find X by conserving nucleon number and proton number. The initial nucleon number is: \(3+12=15\) The initial proton number is: \(2+6=8\) As the alpha particle has a nucleon number of 4 and a proton number of 2 (alpha particle is \({}_{2}^4\mathrm{He}\)), subtract these numbers from initial nucleon and proton numbers to determine X: Nucleon number of X: \(15 - 4 = 11\) Proton number of X: \(8 - 2 = 6\) Therefore, the resulting nucleus X is \({}_{6}^{11}\mathrm{C}\) (carbon with a mass number 11).

c) Calculation of the Q-value of the reaction

The Q-value of the reaction, denoted as \(\mathrm{Q}\), represents the energy released (if positive) or absorbed (if negative) during the reaction. It can be calculated as the difference in the masses of the reactants and the masses of the products, multiplied by the speed of light c^2: \(\mathrm{Q} = (\Delta m) c^2\) Here, \(\Delta m\) represents the mass difference between the reactants and the products. Mass of \({}_{2}^{3}\mathrm{He}\) = 3.01603 u Mass of \({}_{6}^{12}\mathrm{C}\) = 12.00000 u Add both for mass of reactants Mass of \({}_{6}^{11}\mathrm{C}\) = 11.01143 u Mass of alpha particle ( \({}_{2}^4\mathrm{He}\)) = 4.00151 u Add both for mass of products Next, find the mass difference, \(\Delta m\): \(\Delta m = \textrm{mass of reactants} - \textrm{mass of products}\) \(\Delta m = (3.01603 + 12.00000) - (11.01143 + 4.00151)\) \(\Delta m = 0.00309 \textrm{ u}\) Finally, calculate Q using the energy conversion constant for atomic mass units to MeV, which is 931.5 MeV/c²: \(\mathrm{Q} = (\Delta m)(c^2) = (0.00309 \textrm{ u})(931.5 \textrm{ MeV/c}^2)\) \(\mathrm{Q} \approx 2.88 \textrm{ MeV}\)

d) Determining endothermic or exothermic reaction

Since the Q-value of the reaction is positive, it indicates that the reaction releases energy, and therefore, it is an exothermic reaction.

Key concepts

Pick-up Nuclear Reaction

A pick-up nuclear reaction is an intriguing type of reaction where a projectile nucleus swoops in to snatch one or more nucleons from a target nucleus. In this exercise, imagine helium, i.e., \({}_{2}^{3} \mathrm{He}\), launching towards carbon, i.e., \({}_{6}^{12}\mathrm{C}\). This gallant helium nucleus is the projectile, which effectively picks up nucleons from the stationary carbon target. Such reactions are called "pick-up" because the projectile doesn't tackle the target aimlessly; instead, it intentionally picks or captures nucleons from it. The origin of these nucleons is the target nucleus carbon. Therefore, the resulting product is a transformed target nucleus accompanied by the ejected particles, like an alpha particle in this example. Thus, in this nuclear juggling, a new element, commonly called nucleus X, emerges, showing heavy atomic rearrangements and leaving traces like \(\alpha\)-particles in its wake.

Nucleon Conservation

When dealing with nuclear reactions, it's crucial to apply the principle of nucleon conservation. This principle ensures both the nucleon (mass) number and the proton (atomic) number are kept intact throughout the reaction. To get the resulting nucleus X in this reaction, let’s look at the conservation rules:- **Initial nucleon count:** Sum together all nucleons in initial nuclei: \(3 + 12 = 15\).- **Initial proton (atomic) number:** Add up protons from the starters: \(2 + 6 = 8\).An alpha particle \(({}_{2}^{4}\mathrm{He})\) is expelled, contributing - **Nucleons:** 4 - **Protons:** 2Subtract these from the initial calculators:- **Nucleon count for X:** \(15 - 4 = 11\)- **Proton number for X:** \(8 - 2 = 6\)So, the resulting nucleus X becomes \({}_{6}^{11}\mathrm{C}\), signifying a mass number of 11 while retaining the carbon element's attributes.

Q-value Calculation

In nuclear reactions, the Q-value calculation is pivotal as it reflects the energy exchange. A positive Q-value implies energy is released, whereas a negative one means energy is consumed. Here’s how to decode it:1. **Determine Masses of Reactants and Products:** - Mass of \({}_{2}^{3}\mathrm{He}\) = 3.01603 u - Mass of \({}_{6}^{12}\mathrm{C}\) = 12.00000 u - Combined mass of products: - \({}_{6}^{11}\mathrm{C}\) mass = 11.01143 u - \({}_{2}^{4}\mathrm{He}\) (alpha) mass = 4.00151 u2. **Calculate Mass Difference \((\Delta m)\):** \[ \Delta m = (3.01603 + 12.00000) - (11.01143 + 4.00151) = 0.00309 \text{ u} \]3. **Convert Mass Difference to Energy:** Using energy conversion for atomic mass units to MeV: 931.5 MeV/c². \[ \mathrm{Q} = (0.00309 \text{ u})(931.5 \text{ MeV/c}^2) \approx 2.88 \text{ MeV} \]This provides the energy result showing whether the reaction offers energy to the environment or requires it.

Endothermic and Exothermic Reactions

Whether a reaction is endothermic or exothermic can be discerned by its Q-value. The Q-value offers insights into the thermal aspects. Here, since the Q-value is positive, clocking in at approximately 2.88 MeV, it indicates an exothermic reaction. In simple terms: - **Exothermic Reactions:** - Energy release to surroundings. - Often contribute to processes like energy generation in stars. - **Endothermic Reactions:** - Absorbs energy from surroundings. - Usually involve reactions needing energy input to proceed. This Q-value revelation implies the nuclear transformation liberates energy, further validating its impactful nature in nuclear studies.