What Is The Approximate Size Of A Nucleus

How have our ideas almost atoms changed over fourth dimension? - OCR 21C

The diminutive model consists of a nucleus containing protons and neutrons, surrounded past electrons in shells. The numbers of particles in an atom can be calculated from its atomic number and mass number.

i
ii
3
iv
v
6
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Comparing the size and scale of particles

To find out how much bigger one object is than another, divide the larger value by the smaller.

Example:

Two cakes, one with a 40 cm diameter, one with a 20 cm diameter.

Larger value ÷ smaller value

= 40 ÷ 20

= 2

Comparing sizes using society of magnitude

If ane number is 10 times bigger, and so the sizes differ by an guild of magnitude .

Example

The bore of a hydrogen atom is 2.50 × 10 ^-11 yard and the diameter of a gold atom is 1.44 × 10 ^-10 m.

Compare the size of a gold atom with a hydrogen atom. Do their sizes differ by an club of magnitude?

\[\frac{larger~value}{smaller~value} = \frac{one.44 \times 10^{-10}}{two.l \times 10^{-11}} = v.76\]

Therefore a gilded cantlet is five.76 times larger than a hydrogen atom. This is less than 10, so the size of a golden atom is not an social club of magnitude different from the size of a hydrogen atom.

Comparing the size and scale of atoms to the real world

Some objects in the everyday earth are very large. It is easier to write their size in standard grade likewise. For instance, the bore of the Globe is about 13,000,000 m. This is 1.3 × 10 ⁷ m in standard form. To compare the size of atoms to objects in the everyday earth, follow the same method of dividing the larger value by the smaller.

Instance

The diameter of a nucleus is nearly 2 × 10 ^-15 m and the diameter of an atom is one × x ^-10 m. What size would the atom be in a model where the Earth represented the nucleus? The bore of the Earth is 1.iii × 10 ⁷ m.

\[\frac{larger~value}{smaller~value} = \frac{1 \times ten^{-x}}{2 \times x^{-15}} = 5 \times x^4\]

Therefore the atom is 5 × ten ⁴ larger than the nucleus. The model of the cantlet must be v × 10 ⁴ times larger than this.

diameter of the model atom

= (diameter of the Earth) × v × ten ⁴

= 6.v × 10 ¹¹ chiliad

Question

If the World represented an cantlet, what size would the nucleus be? Assume that the diameter of an cantlet is 5 × ten ^four greater than the diameter of the nucleus.

bore of model nucleus = \(\frac{diameter~of~World}{five \times 10^4}\)

\[\frac{i.3 \times ten^seven}{5 \times 10^four}\]

= 260 m