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AlekseyPX
1 year ago
13

“I know what the atomic number of this atom is, but I don’t know what the number of electrons is,” a friend says. How would you

respond?
Physics
1 answer:
liq [111]1 year ago
8 0

Once the atomic number of an atom is known, the number of electrons can be deduced depending on if the atom is an ion or a neutral one.

<h3>Atomic number</h3>

The atomic number of an atom is the number of protons in the nucleus of the atom.

For atoms that are neutral, that is, no net charges, the number of protons is always equal to the number of electrons. In other words, the positive charges always balance the negative charges in neutral atoms.

Thus, if the atomic number of a neutral atom is 6, for example, the proton number will also be 6. Since the proton must balance the electron, the number of electrons will also be 6.

More on atomic numbers can be found here; brainly.com/question/17274608

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The ration of the rms speed of 235uf6 to that of 238uf6 is 1.004.

The molecular mass of 235uf6 is 349, while that of 238uf6 is 352.

The rms speed is calculated as

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Thus the ratio rms speed of 235uf6 to 238uf6 is calculated as

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An electron is fired horizontally at 2.5 x 10 m/s between two horizontal parallel plates 7.5 cm long, as shown in the diagram. T
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Explanation:

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2 years ago
What is the term for producing a current by moving a wire through a magnetic field?.
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The answer is electromagnetic Induction.



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2 years ago
A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
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Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = \frac{I}{nqA}

Where;

n = number of free electrons per cubic meter

q =  electron charge

A =  cross-sectional area of the wire

<em>First let's calculate the number of free electrons per cubic meter (n)</em>

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M)          -------------(ii)

<em>Substitute the values of Nₐ, ρ and M into equation (ii) as follows;</em>

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

<em>Now let's calculate the drift electron</em>

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

<em>Substitute these values into equation (i) as follows;</em>

v = \frac{I}{nqA}

v = \frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

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