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Chapter 5: Problem 179

With the advent of techniques such as scanning tunneling microscopy, it is now possible to "write" with individual atoms by manipulating and arranging atoms on an atomic surface. a. If an image is prepared by manipulating iron atoms and their total mass is \(1.05 \times 10^{-20} \mathrm{g},\) what number of iron atoms were used? b. If the image is prepared on a platinum surface that is exactly 20 platinum atoms high and 14 platinum atoms wide, what is the mass (grams) of the atomic surface? c. If the atomic surface were changed to ruthenium atoms and the same surface mass as determined in part b is used, what number of ruthenium atoms is needed to construct the surface?

Short Answer

Expert verified
a. There were \(1.12\times10^4\) iron atoms used in the image. b. The mass of the platinum atomic surface is approximately \(8.98 \times 10^{-21}\mathrm{g}\). c. \(5.42\times 10^2\) ruthenium atoms are needed to construct the surface with the same mass as the platinum surface.

Step by step solution

01

Find the number of iron atoms used

To find the number of iron atoms used, we need to divide the mass of the iron atoms by the atomic mass of iron. The atomic mass of iron is approximately 56 u (unified atomic mass units), where 1 u = 1.6605 × 10^-24 g. So, we have: Number of Iron atoms = (Total mass of iron atoms) / (Atomic mass of iron in grams) \( Number\, of\, Iron\, atoms=\frac{1.05\times 10^{-20}\, g}{(56\, u)\times(1.6605 \times 10^{-24} \mathrm{g/u})} \)
02

Calculate the number of iron atoms

Now we can calculate the number of iron atoms: \( Number\, of\, Iron\, atoms=\frac{1.05\times 10^{-20}\, g}{(56\, u)\times(1.6605 \times 10^{-24}\, g/u)}=1.12\times10^4\, atoms \) So, there were 1.12 x 10^4 iron atoms used.
03

Calculate the mass of the platinum surface

To find the mass of the atomic surface made of platinum, we first need to find the number of platinum atoms in the surface. Since it is 20 platinum atoms high and 14 platinum atoms wide, there are 20 x 14 = 280 platinum atoms. The atomic mass of platinum is about 195 u. Mass of the platinum surface = (Number of platinum atoms) x (Atomic mass of platinum in grams) \( Mass\, of\, the\, platinum\, surface=280\, atoms\, \times\, (195\, u)\times(1.6605 \times 10^{-24} \mathrm{g/u}) \)
04

Calculate the mass of the platinum surface

Now we can calculate the mass of the platinum surface: \( Mass\, of\, the\, platinum\, surface=280\, atoms\, \times\, (195\, u)\times(1.6605 \times 10^{-24}\, g/u)=8.9831\times10^{-21}\, g \) So, the mass of the platinum atomic surface is approximately 8.98 × 10^-21 g.
05

Find the number of ruthenium atoms needed

To find the number of ruthenium atoms needed, we need to divide the mass of the platinum surface by the atomic mass of ruthenium. The atomic mass of ruthenium is approximately 101 u. Number of ruthenium atoms = (Mass of platinum surface) / (Atomic mass of ruthenium in grams) \( Number\, of\, ruthenium\, atoms=\frac{8.9831\times 10^{-21}\, g}{(101\, u)\times(1.6605 \times 10^{-24} \mathrm{g/u})} \)
06

Calculate the number of ruthenium atoms

Now we can calculate the number of ruthenium atoms: \( Number\, of\, ruthenium\, atoms=\frac{8.9831\times 10^{-21}\, g}{(101\, u)\times(1.6605 \times 10^{-24}\, g/u)}=5.42\times 10^2\, atoms \) So, 5.42 x 10^2 ruthenium atoms are needed to construct the surface with the same mass as the platinum surface.

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