Home Page Textbooks Physics: Principles and Problems Page 778: Practice Problems

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

Textbook solutions

All Solutions

Page 778: Practice Problems

Exercise 1

Step 1

1 of 4

In this problem, we are given the chemical element zinc with known density and atomic mass and number of free electrons per atom. We want to find the number of free electrons in one cubic centimeter of zinc.

Step 2

2 of 4

The quantity we are looking for is simply the density of free electrons and we can write it as the product of the number of zinc atoms in the same volume and the factor two. The former is expressed via Avogadro’s number, molar mass, and density so we have that

$$
rho_{free}=frac{2times N_A rho}{M}
$$

Step 3

3 of 4

Finally, we have that the number of the free electrons is obtained after plugging the values as

$$
rho_{free}=frac{2times 6.023times 10^{23}times 7.13}{65.37}=boxed{1.31times 10^{23}frac{textrm{e}^-}{textrm{cm}^{3}}}
$$

Result

4 of 4

$$
rho_{free}=1.31times 10^{23}frac{textrm{e}^-}{textrm{cm}^{3}}
$$

Exercise 2

Step 1

1 of 4

In this problem, we are given the chemical element silver with known density and atomic mass, and the number of free electrons per atom. We want to find the number of free electrons in one cubic centimeter of silver.

Step 2

2 of 4

The quantity we are looking for is simply the density of free electrons and we can write it as the product of the number of silver atoms in the same volume and the factor one. The former is expressed via Avogadro’s number, molar mass, and density so we have that

$$
rho_{free}=frac{rho N_A }{M}
$$

Step 3

3 of 4

Finally, we have that the number of the free electrons is obtained after plugging the values as

$$
rho_{free}=frac{6.023times 10^{23}times 10.49}{107.87}=boxed{0.59times 10^{23}frac{textrm{e}^-}{textrm{cm}^{3}}}
$$

Result

4 of 4

$$
rho_{free}=0.59times 10^{23}frac{textrm{e}^-}{textrm{cm}^{3}}
$$

Exercise 3

Step 1

1 of 4

In this problem, we are given the chemical element gold with known density and atomic mass, and the number of free electrons per atom. We want to find the number of free electrons in one cubic centimeter of gold.

Step 2

2 of 4

The quantity we are looking for is simply the density of free electrons and we can write it as the product of the number of gold atoms in the same volume and the factor one. The former is expressed via Avogadro’s number, molar mass, and density so we have that

$$
rho_{free}=1timesfrac{rho N_A }{M}
$$

Step 3

3 of 4

Finally, we have that the number of the free electrons is obtained after plugging the values as

$$
rho_{free}=frac{6.022times 10^{23}times 19.32}{196.97}=boxed{0.59times 10^{23}frac{textrm{e}^-}{textrm{cm}^{3}}}
$$

Result

4 of 4

$$
rho_{free}=0.59times 10^{23}frac{textrm{e}^-}{textrm{cm}^{3}}
$$

Exercise 4

Step 1

1 of 4

In this problem, we are given the chemical element aluminum with known density and atomic mass, and the number of free electrons per atom. We want to find the number of free electrons in one cubic centimeter of aluminum.

Step 2

2 of 4

The quantity we are looking for is simply the density of free electrons and we can write it as the product of the number of aluminum atoms in the same volume and the factor three. The former is expressed via Avogadro’s number, molar mass, and density so we have that

$$
rho_{free}=3timesfrac{rho N_A }{M}
$$

Step 3

3 of 4

Finally, we have that the number of the free electrons is obtained after plugging the values as

$$
rho_{free}=3timesfrac{6.022times 10^{23}times 2.7}{26.92}=boxed{1.81times 10^{23}frac{textrm{e}^-}{textrm{cm}^{3}}}
$$

Result

4 of 4

$$
rho_{free}=1.81times 10^{23}frac{textrm{e}^-}{textrm{cm}^{3}}
$$

Exercise 5

Step 1

1 of 5

In this problem, we are given the chemical element aluminum with known density and atomic mass, and the number of free electrons per atom. Our task is to calculate how many free electrons do we have in 2835 g of aluminum. We can do so, by first finding the number of free electrons in a cubic centimeter of aluminum and then we find how many cubic centimeters of aluminum are there in the given tip of the monument.

Step 2

2 of 5

First, we have that the number of the free electrons is obtained as

$$
rho_{free}=3timesfrac{6.022times 10^{23}times 2.7}{26.92}=1.81times 10^{23}frac{textrm{e}^-}{textrm{cm}^{3}}
$$

Step 3

3 of 5

The given body of aluminum occupies a volume that is found from the following equation

$$
V=frac{m}{rho}=frac{2835}{2.7}=1050textrm{ cm}^3
$$

Step 4

4 of 5

Now, the total number of the electrons in the tip of the monument is simply given as

$$
N=rho_{free}times V=1.81times 10^{23}times 1050=boxed{1.9times 10^{26}textrm{ e}^-}
$$

Result

5 of 5

$$
N=1.9times 10^{26}textrm{ e}^-
$$

Exercise 6

Step 1

1 of 4

In this problem we are given the chemical element germanium with known density, atomic mass and number of free electrons per cubic centimeter. We want to know the number of free electrons per atom.

Step 2

2 of 4

In order to do so, we are going to start from the formula that tells us how many free electrons do we have in a cubic centimeter

$$
rho_{free}=n_{atom}frac{rho N_A}{M}
$$

From where we see that

$$
n_{atom}=frac{rho_{free}M}{rho N_A}
$$

Step 3

3 of 4

Now, we can use the equation obtained above and plug in the given values to have that

$$
n_{atom}=frac{2.25times 10^{13}times 72.6}{5.23times 6.022times 10^{23}}=boxed{5.19times 10^{-10} frac{textrm{e}^-}{textrm{atom}}}
$$

Result

4 of 4

$$
n_{atom}=5.19times 10^{-10} frac{textrm{e}^-}{textrm{atom}}
$$

Exercise 7

Step 1

1 of 5

In this problem, we are given the chemical element silicon with known density, atomic mass, and the number of free electrons per cubic centimeter. We want to know the number of free electrons per atom. After that, we should express the given temperature in degrees Celsius.

Step 2

2 of 5

In order to do so, we are going to start from the formula that tells us how many free electrons do we have in a cubic centimeter

$$
rho_{free}=n_{atom}frac{rho N_A}{M}
$$

From where we see that

$$
n_{atom}=frac{rho_{free}M}{rho N_A}
$$

Step 3

3 of 5

Now, we can use the equation obtained above and plug in the given values to have that

$$
n_{atom}=frac{1.89times 10^{5}times 28.09}{2.33times 6.022times 10^{23}}=boxed{3.8times 10^{-18} frac{textrm{e}^-}{textrm{atom}}}
$$

Step 4

4 of 5

The temperature given in Kelvins can be expressed in degrees Celsius in a very simple manner

$$
T_C=T_K-273=200-273=boxed{-73^circ textrm{ C}}
$$

Result

5 of 5

$$
n_{atom}=3.8times 10^{-18} frac{textrm{e}^-}{textrm{atom}}
$$

$$
T_C=-73^circ textrm{ C}
$$

Exercise 8

Step 1

1 of 5

Step 2

2 of 5

In order to do so, we are going to start from the formula that tells us how many free electrons do we have in a cubic centimeter

$$
rho_{free}=n_{atom}frac{rho N_A}{M}
$$

From where we see that

$$
n_{atom}=frac{rho_{free}M}{rho N_A}
$$

Step 3

3 of 5

Now, we can use the equation obtained above and plug in the given values to have that

$$
n_{atom}=frac{9.23times 10^{-10}times 28.09}{2.33times 6.022times 10^{23}}=boxed{1.8times 10^{-32} frac{textrm{e}^-}{textrm{atom}}}
$$

Step 4

4 of 5

The temperature given in Kelvins can be expressed in degrees Celsius in a very simple manner

$$
T_C=T_K-273=100-273=boxed{-173^circ textrm{ C}}
$$

Result

5 of 5

$$
n_{atom}=1.8times 10^{-32} frac{textrm{e}^-}{textrm{atom}}
$$

$$
T_C=-173^circ textrm{ C}
$$

Exercise 9

Step 1

1 of 4

In this problem we are given the chemical element germanium with known density, atomic mass and number of free electrons per cubic centimeter. We want to know the number of free electrons per atom.

Step 2

2 of 4

In order to do so, we are going to start from the formula that tells us how many free electrons do we have in a cubic centimeter

$$
rho_{free}=n_{atom}frac{rho N_A}{M}
$$

From where we see that

$$
n_{atom}=frac{rho_{free}M}{rho N_A}
$$

Step 3

3 of 4

Now, we can use the equation obtained above and plug in the given values to have that

$$
n_{atom}=frac{1.16times 10^{10}times 72.6}{5.23times 6.022times 10^{23}}=boxed{2.7times 10^{-13} frac{textrm{e}^-}{textrm{atom}}}
$$

Result

4 of 4

$$
n_{atom}=2.7times 10^{-13} frac{textrm{e}^-}{textrm{atom}}
$$

Exercise 10

Step 1

1 of 4

In this problem we are given the chemical element germanium with known density, atomic mass and number of free electrons per cubic centimeter. We want to know the number of free electrons per atom.

Step 2

2 of 4

In order to do so, we are going to start from the formula that tells us how many free electrons do we have in a cubic centimeter

$$
rho_{free}=n_{atom}frac{rho N_A}{M}
$$

From where we see that

$$
n_{atom}=frac{rho_{free}M}{rho N_A}
$$

Step 3

3 of 4

Now, we can use the equation obtained above and plug in the given values to have that

$$
n_{atom}=frac{3.47times 72.6}{5.23times 6.022times 10^{23}}=boxed{8times 10^{-23} frac{textrm{e}^-}{textrm{atom}}}
$$

Result

4 of 4

$$
n_{atom}=8times 10^{-23} frac{textrm{e}^-}{textrm{atom}}
$$

Exercise 11

Step 1

1 of 4

In this problem, we are given aresenic-doped silicon for which we should aim to have 10,000 electrons coming from the arsenic. We should find the ratio of arsenic and silicon.

Step 2

2 of 4

Since we know that there is one free electron per arsenic atom the ratio we are looking for is going to be given as

$$
frac{N_{As}}{N_{Si}}=10^4times frac{N_{freeSi}}{N_{Si}}
$$

where $N_{freeSi}$ and $N_{Si}$ represent the number of silicon free electrons and atoms per a cubic centimeter, respectively.

Step 3

3 of 4

Referring to known parametric values of the silicon and plugging them into the above expression we obtain that

$$
frac{N_{As}}{N_{Si}}=10^4times frac{1.45times 10^{10}}{4.99times 10^{22}}=boxed{2.9times 10^{-9}}
$$

Result

4 of 4

$$
frac{N_{As}}{N_{Si}}=2.9times 10^{-9}
$$

Exercise 12

Step 1

1 of 5

In this problem, we are given arsenic-doped germanium for which we should aim to have 5000 times more electrons coming from the arsenic. We should find the ratio of arsenic and silicon.

Step 2

2 of 5

Since we know that there is one free electron per arsenic atom the ratio we are looking for is going to be given as

$$
frac{N_{As}}{N_{Ge}}=5times10^3times frac{N_{freeGe}}{N_{Ge}}=5times10^3 times n_{atom}
$$

where $N_{freeGe}$ and $N_{Ge}$ represent the number of germanium free electrons and atoms per cubic centimeter, respectively which give a ratio of the free electrons per atom.

Step 3

3 of 5

center{The number of free electrons per atom in germanium can be found as }
[n_{atom}=frac{rho_{free}M}{rho N_A}=frac{2.25times 10^{13}times 72.6}{5.23times 6.022times 10^{23}}=5.19times 10^{-10} frac{textrm{e}^-}{textrm{atom}}]

Step 4

4 of 5

Referring to known parametric values of the germanium,and plugging them into the above expression we obtain that

$$
frac{N_{As}}{N_{Ge}}=5times 10^3times 5.19times 10^{-10}=boxed{2.6times 10^{-6}}
$$

Result

5 of 5

$$
frac{N_{As}}{N_{Ge}}=2.6times 10^{-6}
$$

Exercise 13

Step 1

1 of 6

In this problem we are given germanium which is arsenic-doped with the ratio 1:1,000,000. We want to know the ratio between the thermal and doped carriers.

Step 2

2 of 6

Since we know that there is one free electron per arsenic atomwe have that

$$
frac{N_{As}}{N_{Ge}}=frac{N_{freeAs}}{N_{freeGe}}times frac{N_{freeGe}}{N_{Ge}}=frac{N_{freeGe}}{N_{freeAs}}times n_{atom}
$$

where $N_{freeGe}$ and $N_{Ge}$ represent the number of germanium free electrons and atoms per cubic centimeter, respectively which give a ratio of the free electrons per atom.

Step 3

3 of 6

So we have that the ratio we are looking for can be written as

$$
frac{N_{freeAs}}{N_{freeGe}}=frac{N_{Ge}}{N_{As}}times frac{1}{n_{atom}}
$$

Step 4

4 of 6

center{The number of free electrons per atom in germanium can be found as }
[n_{atom}=frac{rho_{free}M}{rho N_A}=frac{1.13times 10^{15}times 72.6}{5.23times 6.022times 10^{23}}=2.62times 10^{-8} frac{textrm{e}^-}{textrm{atom}}]

Step 5

5 of 6

center{Finally, we can write that }
[frac{N_{freeAs}}{N_{freeGe}}=frac{1}{10^6}times frac{1}{2.62times 10^{-10}}=boxed{38}]

Result

6 of 6

$$
frac{N_{freeAs}}{N_{freeGe}}=boxed{38}
$$

Exercise 14

Step 1

1 of 6

In this problem, we are given silicon which is arsenic-doped with the ratio of 1:1,000,000. We want to know the ratio between the thermal and doped carriers.

Step 2

2 of 6

Since we know that there is one free electron per arsenic atomwe have that

$$
frac{N_{As}}{N_{Si}}=frac{N_{freeAs}}{N_{freeSi}}times frac{N_{freeSi}}{N_{Si}}=frac{N_{freeSi}}{N_{freeAs}}times n_{atom}
$$

where $N_{freeSi}$ and $N_{Si}$ represent the number of silicon free electrons and atoms per cubic centimeter, respectively which give a ratio of the free electrons per atom.

Step 3

3 of 6

So we have that the ratio we are looking for can be written as

$$
frac{N_{freeAs}}{N_{freeSi}}=frac{N_{Si}}{N_{As}}times frac{1}{n_{atom}}
$$

Step 4

4 of 6

center{The number of free electrons per atom in silicon at given temperature can be found as }
[n_{atom}=frac{rho_{free}M}{rho N_A}=frac{4.54times 10^{12}times 28.09}{2.33times 6.022times 10^{23}}=9.09times 10^{-11} frac{textrm{e}^-}{textrm{atom}}]

Step 5

5 of 6

center{Finally, we can write that }
[frac{N_{freeAs}}{N_{freeGe}}=frac{1}{10^6}times frac{1}{9.09times 10^{-11}}=boxed{11times 10^3}]

Result

6 of 6

$$
frac{N_{freeAs}}{N_{freeGe}}=11times 10^3
$$

Exercise 15

Step 1

1 of 1

Germanium has way too many thermally liberated carriers relative to the doped ones and therefore the conductivity will start to decrease. This is why the silicon is proffered material at higher temperatures.

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Page 778: Practice Problems

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