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Chapter 2: Problem 7

The diameter of a neutral helium atom is about \(1 \times\) \(10^{2} \mathrm{pm} .\) Suppose that we could line up helium atoms side by side in contact with one another. Approximately how many atoms would it take to make the distance from end to end \(1 \mathrm{~cm} ?\)

Short Answer

Expert verified
It takes approximately \(1 \times 10^8\) helium atoms to span 1 cm.

Step by step solution

01

Convert cm to pm

Start by converting the length from centimeters to picometers, as the diameter of an atom is given in picometers (pm). Note that 1 cm equals \(10^{10}\) pm. Therefore, converting 1 cm to pm results in \(1 \times 10^{10}\) pm.
02

Calculate the number of helium atoms

The diameter of one helium atom is approximately \(1 \times 10^2\) pm. To find out how many atoms would fit side by side in a length of \(1 \times 10^{10}\) pm, divide the total length by the diameter of one atom: \( \frac{1 \times 10^{10} \text{ pm}}{1 \times 10^2 \text{ pm/atom}} = 1 \times 10^8 \text{ atoms} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Metric Conversion
When dealing with measurements on an atomic scale, it is crucial to understand how metric conversions work. For example, converting between units like centimeters (cm) and picometers (pm) can seem daunting, but it becomes manageable once you know the basic conversion factors.

The metric system is based on powers of ten. This makes conversion straightforward, as you simply need to adjust the exponent based on the unit scale differences:
  • 1 centimeter (cm) is equivalent to 10 millimeters (mm).
  • 1 millimeter is equal to 1000 micrometers (\(\mu m\)).
  • 1 micrometer is the same as 1000 nanometers (nm).
  • 1 nanometer translates to 10,000 picometers (pm).
Knowing these conversions helps us see that 1 cm equals \(1 \times 10^{10}\) pm. In scenarios where precision is necessary, converting cm to pm is vital, especially when dealing with tiny atomic distances.
Calculation
Performing calculations with atoms requires a basic understanding of how to handle large numbers efficiently, especially when powers of ten are involved. Calculating how many helium atoms fit in a certain length involves just a few steps, but understanding each one is key.

After converting 1 cm to picometers, you have \(1 \times 10^{10}\) pm as the total length. Knowing that the diameter of a helium atom is \(1 \times 10^2\) pm, the process is as follows:
  • You'll divide the total length in picometers by the diameter of one atom in picometers.
  • The formula is simple: \(\frac{1 \times 10^{10} \text{ pm}}{1 \times 10^2 \text{ pm/atom}}\).
  • When dividing powers of ten, you subtract the exponents: \(10^{10} - 10^2 = 10^8\).
  • The result is \(1 \times 10^8\), meaning 100 million helium atoms can fit side by side along 1 cm.
Understanding these calculation steps helps with any similar problem involving linear arrangements of atoms.
Atomic Diameter
Atomic diameter is a fundamental concept in atomic structure and refers to the distance across an atom. The diameter is often extremely tiny and measured in picometers (pm), which are a billionth of a millimeter.

It's crucial to grasp this concept because it provides insight into the scale of atoms and their interactions.
  • Helium atoms have a diameter of about \(1 \times 10^2\) pm, which indicates their small size.
  • Understanding atomic diameters helps in fields such as nanotechnology and chemistry, where precise manipulations of matter are essential.
  • Atomic diameter influences properties such as the ability of atoms to combine to form molecules and their behavior in materials.
By appreciating the small yet significant atomic diameter, scientists can better predict how materials will behave and develop new technologies at the molecular level.

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Most popular questions from this chapter

Calculate the number of neutrons of \({ }^{239} \mathrm{Pu}\).

Which of the following compounds are likely to be ionic? Which are likely to be molecular? \(\mathrm{CH}_{4},\) NaBr, \(\mathrm{BaF}_{2}, \mathrm{CCl}_{4}, \mathrm{ICl}, \mathrm{CsCl}, \mathrm{NF}_{3}\)

While most isotopes of light elements such as oxygen and phosphorus contain relatively equal amounts of protons and neutrons in the nucleus, recent results indicate that a new class of isotopes called neutronrich isotopes can be prepared. These neutron-rich isotopes push the limits of nuclear stability as the large numbers of neutrons approach the "neutron drip line." These neutron-rich isotopes may play a critical role in the nuclear reactions of stars. Determine the number of neutrons in the following neutron-rich isotopes: (a) \({ }^{40} \mathrm{Mg},(\mathrm{b}){ }^{44} \mathrm{Si},(\mathrm{c}){ }^{48} \mathrm{Ca},(\mathrm{d}){ }^{43} \mathrm{Al}\)

(a) Which elements are most likely to form ionic compounds? (b) Which metallic elements are most likely to form cations with different charges?

Fill the blanks in the following table. $$ \begin{array}{|c|c|c|c|} \hline \text { Cation } & \text { Anion } & \text { Formula } & \text { Name } \\\ \hline & & & \text { Magnesium bicarbonate } \\ \hline & & \mathrm{SrCl}_{2} & \\ \hline \mathrm{Fe}^{3+} & \mathrm{NO}_{2}^{-} & & \\ \hline & & & \text { Manganese(II) chlorate } \\ \hline & & \mathrm{SnBr}_{4} & \\ \hline \mathrm{Co}^{2+} & \mathrm{PO}_{4}^{3-} & & \\ \hline \mathrm{Hg}_{2}^{2+} & \mathrm{I}^{-} & & \\ \hline & & \mathrm{Cu}_{2} \mathrm{CO}_{3} & \\ \hline & & & \text { Lithium nitride } \\ \hline \mathrm{Al}^{3+} & \mathrm{S}^{2-} & & \\ \hline \end{array} $$
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