Question

#4 chemistry a schoology 2025

1. 20 protons and 24 neutrons
2. 40 protons and 20 neutrons
3. 60 protons and 80 neutrons
4. atomic number = 40, mass number = 70
5. atomic number = 80, mass number = 180

:: stable :: unstable

Explanation:

Step1: Recall neutron - proton ratio concept

For light elements (low number of protons), a stable nucleus has a neutron - proton ($n/p$) ratio close to 1. For heavier elements, the stable $n/p$ ratio is greater than 1.

Step2: Calculate $n/p$ ratio for 1

Number of protons ($Z$) = 20, number of neutrons ($N$)=24. $n/p=\frac{24}{20} = 1.2$. Looking at the graph, for $Z = 20$, this ratio is within the stable region. So it is stable.

Step3: Calculate $n/p$ ratio for 2

Number of protons ($Z$) = 40, number of neutrons ($N$)=20. $n/p=\frac{20}{40}=0.5$. For $Z = 40$, this ratio is outside the stable region. So it is unstable.

Step4: Calculate $n/p$ ratio for 3

Number of protons ($Z$) = 60, number of neutrons ($N$)=80. $n/p=\frac{80}{60}=\frac{4}{3}\approx1.33$. For $Z = 60$, this ratio is within the stable region. So it is stable.

Step5: Calculate number of neutrons and $n/p$ ratio for 4

Atomic number ($Z$)=40, Mass number ($A$) = 70. Number of neutrons ($N$)=A - Z=70 - 40 = 30. $n/p=\frac{30}{40}=0.75$. For $Z = 40$, this ratio is outside the stable region. So it is unstable.

Step6: Calculate number of neutrons and $n/p$ ratio for 5

Atomic number ($Z$)=80, Mass number ($A$) = 180. Number of neutrons ($N$)=A - Z=180 - 80 = 100. $n/p=\frac{100}{80}=1.25$. For $Z = 80$, this ratio is outside the stable region. So it is unstable.

Answer:

1. Stable
2. Unstable
3. Stable
4. Unstable
5. Unstable