1. The atomic number deceases by one and the atomic mass is unchanged - β⁺/positron emission. In this a proton is converted to a neutron, hence no net change in mass. Since a proton is converted to neutron, the daughter nuclide's proton number /atomic number decreases by 1. Then atomic number decreases by 1 and no change in mass.
2.the atomic number decreases by two, and the atomic mass decreases by four - alpha decay / ⁴₂α. alpha particles have 2 neutrons and 2 protons. Since 2 protons and 2 neutrons are emitted, the mass of the nuclide decreases by 4. Since 2 protons have been emitted, this results in atomic number decreasing by 2.
3.<span>the atomic number increases by one, and the atomic mass remains unchanged. - </span>β⁻ beta decay. In this type of decay - beta decay , a neutron is converted to a proton, therefore no net change in mass. Since a proton is formed, atomic number increases by 1. Therefore atomic number increases by 1 and no change in mass.
Answer:
Here is ur answer;
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Answer:
X = Water (H2O) ; Y = Hydrogen ; Z = Oxygen
Explanation:
2(H2O) -------> 2H2 + O2
The atomic number of Li is 3
Electron configuration of Li : 1s² 2s¹
The atomic number of Na is 11
Electron configuration of Na : 1s²2s²2p⁶3s¹
Thus there is one electron in the valence shell of Li (2s¹) and that of Na (3s¹). However, the valence electron in Na is in a shell that is farther away from the nucleus compared to that of Li. As a result, the Na valence electron will be held less tightly by the nucleus i.e. it will experience a reduced nuclear attraction and can be removed easily than the Li 2s electron.
I think the answer is D in my opinion
