Compléter les équations stœchiométriques suivantes : a) 26^58 Fe+2 0^1 n= 27^60 Co+……b) 20^40

Compléter les équations stœchiométriques suivantes : a) ...

Compléter les équations stœchiométriques suivantes : a) ${ }_{26}^{58} \mathrm{Fe}+2{ }_0^1 \mathrm{n}={ }_{27}^{60} \mathrm{Co}+\ldots \ldots$ b) ${ }_{20}^{40} \mathrm{Ca}+{ }_1^2 \mathrm{~d}=\ldots \ldots+{ }_1^1 \mathrm{p}$ c) ${ }_{28}^{60} \mathrm{Ni}+\ldots \ldots={ }_{27}^{60} \mathrm{Co}+{ }_1^1 \mathrm{p}$ d) ${ }_{96}^{242} \mathrm{Cm}+{ }_2^4 \alpha=\ldots \ldots+{ }_0^1 \mathrm{n}$ e) ${ }_{15}^{30} \mathrm{P}=\ldots \ldots+{ }_{+1}^0 \mathrm{e}$ f) ${ }_{92}^{235} \mathrm{U}+{ }_0^1 \mathrm{n}={ }_{56}^{142} \mathrm{Ba}+\ldots \ldots+2{ }_0^1 \mathrm{n}$ g) ${ }^{35} \mathrm{Cl}+\mathrm{n}={ }^{35} \mathrm{~S}+\ldots \ldots$ h) ${ }^{209} \mathrm{Bi}+\ldots \ldots .={ }^{210} \mathrm{Bi}+\mathrm{p}$ i) ${ }^{58} \mathrm{Fe}+\mathrm{n}={ }^{59} \mathrm{Co}+\ldots \ldots$ j) ${ }^{59} \mathrm{Co}+\ldots \ldots .={ }^{56} \mathrm{Mn}+\alpha$ k) ${ }^{214} \mathrm{~Pb}={ }^{214} \mathrm{Bi}+\ldots \ldots$.

Compléter les équations stœchiométriques suivantes :
a) ${ }_{26}^{58} \mathrm{Fe}+2{ }_0^1 \mathrm{n}={ }_{27}^{60} \mathrm{Co}+\ldots \ldots$
b) ${ }_{20}^{40} \mathrm{Ca}+{ }_1^2 \mathrm{~d}=\ldots \ldots+{ }_1^1 \mathrm{p}$
c) ${ }_{28}^{60} \mathrm{Ni}+\ldots \ldots={ }_{27}^{60} \mathrm{Co}+{ }_1^1 \mathrm{p}$
d) ${ }_{96}^{242} \mathrm{Cm}+{ }_2^4 \alpha=\ldots \ldots+{ }_0^1 \mathrm{n}$
e) ${ }_{15}^{30} \mathrm{P}=\ldots \ldots+{ }_{+1}^0 \mathrm{e}$
f) ${ }_{92}^{235} \mathrm{U}+{ }_0^1 \mathrm{n}={ }_{56}^{142} \mathrm{Ba}+\ldots \ldots+2{ }_0^1 \mathrm{n}$
g) ${ }^{35} \mathrm{Cl}+\mathrm{n}={ }^{35} \mathrm{~S}+\ldots \ldots$
h) ${ }^{209} \mathrm{Bi}+\ldots \ldots .={ }^{210} \mathrm{Bi}+\mathrm{p}$
i) ${ }^{58} \mathrm{Fe}+\mathrm{n}={ }^{59} \mathrm{Co}+\ldots \ldots$
j) ${ }^{59} \mathrm{Co}+\ldots \ldots .={ }^{56} \mathrm{Mn}+\alpha$
k) ${ }^{214} \mathrm{~Pb}={ }^{214} \mathrm{Bi}+\ldots \ldots$.

Key Concepts

Nuclear Decay Modes

Nuclear decay modes, such as alpha decay and beta decay, describe the processes through which unstable nuclei achieve stability. Alpha decay involves the emission of a helium nucleus while beta decay typically involves the emission of an electron or a positron. Recognizing these decay patterns is essential for predicting and balancing the observed nuclear transformations.

Types of Nuclear Processes

Nuclear reactions encompass a variety of processes, including neutron capture, proton capture or emission, alpha decay, beta decay, and fission processes. Each of these involves distinct changes within the nucleus and the emission or absorption of specific particles, which are crucial for understanding how different nuclear reactions proceed.

Conservation of Mass and Atomic Numbers

In nuclear reactions the total mass number (sum of protons and neutrons) and the atomic number (number of protons) are conserved. This means that when writing a nuclear equation, the sum of these numbers on the reactants side must equal the corresponding sum on the products side. This principle is fundamental for balancing nuclear reactions.

Balancing Nuclear Equations

Balancing nuclear equations requires the identification and addition of missing nuclear particles by ensuring that both the mass numbers and atomic numbers remain equal on both sides of the reaction. This methodical approach is analogous to balancing chemical equations and is central to accurately representing nuclear transformations.

Transcript

00:04 In order to find an unknown particle of a nuclear reaction, we start by adding the mass numbers of our reactants and setting them equal to the sum of the mass numbers of our products.

00:16 Then we add the atomic numbers of our reactants and set them equal to the sum of the atomic numbers of our products.

00:25 Then we look at the atomic number of our unknown particle and identify it from the periodic table.

00:31 So starting with our first reaction.

00:35 On the left -hand side, we have 54 plus 4 is 58 for our mass numbers.

00:40 And on the right -hand side, we have 2 times 1 is 2 plus some number is 58.

00:47 So 2 plus 56 is 58.

00:51 For our atomic numbers of our reactants, on the left -hand side, we have 26 plus 2 is 28th.

00:58 And on the bottom, we have zero plus some number is 28, so that has to be 28.

01:05 Looking on the periodic table for atomic number 28, we find our unknown particle is nickel 56.

01:17 Next, we have an aluminum 27 being bombarded by an alpha particle and producing phosphorus 30 plus an unknown particle.

01:29 So 27 plus 4 is 31.

01:33 30 plus 1 gives us 31.

01:38 On the bottom, we have 13 plus 2 is 15.

01:42 15 plus 0 gives us 15.

01:46 So our unknown particle is a neutron that's being produced.

01:52 For the next reaction, sulfur being bombarded by a neutron produces a proton.

02:00 Or a hydrogen nuclei.

02:03 And we have 32 plus 1 is 33 on the reactant side.

02:09 1 plus 32 goes on our product side to give us 33...

Please give Ace some feedback