For Al, its atomic number is 13 and its mass number is 27. How many neutrons does it have? A. 13 B. 14 C. 26 D. 27 E. 40

Physics

1 answer:

sveta [45]4 years ago

5 0

Answer:

B. 14

Explanation:

For an element:

- Atomic number: is the number of protons in the nucleus (p)

- Mass number: it is the number of protons (p) + neutrons (n) in the nucleus

For Al (aluminium), we have:

p = 13

p + n = 27

Substituting the first equation into the second one, we find the number of neutrons:

n = 27 - p = 27 - 13 = 14

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<h3>Further explanation</h3>

Matter is the building block of everything that we encounter in our lives. Matter is made of atoms, which are in turn made of tiny particles which are called electrons, protons, and neutrons. The ammount of these 3 elements, and their topological configuration in the atoms, is what determines what a certain element is (like Carbon, Hydrogen, Iron, etc).

In some cases, some elements may lose or gain some electrons. Regarded that this missing or extra electrons are not very high in number, the material doesn't lose any of its properties, however it will always try to get its number of electrons back to normal. This is when we say that an element has a <em>charge</em>, which is a measure of how much electrons a body needs to get back to normal. A body has positive charge if it lacks electrons, and has negative charge if it has extra electrons.

This charge causes the material to have an Electric field, which is a measure of how much does it attract or repel electrons. In the case of our problem, we need to compute exactly that, the Electric field. In our problem, we have an infinitely long cylinder with a linear charge density $\lambda$ , this means that all parts of the cylinder have the same charge, and due to symmetry, the electric field is constant on the angular and longitudinal directions of the cylinder.

This makes easy to apply Gauss' Law, since for a Gaussian curve in the shape of a concentric cylinder (with a higher radius than that of our charged cylinder) we can write:

$\Phi = \frac{\lambda \cdot L}{\epsilon_0}$

Where $\Phi$ is called the Electric flux. Since the electric field is constant for a given distance <em>r</em> from the axis of the cylinder we can write that: