Physics: Principles and Problems
9th Edition
ISBN: 9780078458132
Table of contents
Textbook solutions
All Solutions
Page 759: Section Review
Exercise 9
Step 1
1 of 1
Rutherford’s nuclear model or planetary model resembles the solar system where most of the atom’s mass is located in the center while electrons orbit around it like planets orbit around the Sun.
Exercise 10
Step 1
1 of 2
Two spectra differ in the fact that the one coming from the incandescent solids is a continuous band of colors whereas the one from gases is made of discrete color lines.
Step 2
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They are similar because they both have the same origin, the energy-level transition in atoms.
Exercise 11
Step 1
1 of 1
Upon the photon is absorbed by the atom the final atom energy equals the initial atom energy plus energy of the photon.
$$
E_f=E_i+E_{ph}
$$
Exercise 12
Step 1
1 of 4
In this problem we should use Bohr model to find the radius of the second energy level in a helium ion.
Step 2
2 of 4
To do so, we will use the fact that in Bohr model the radius of each energy level is
$$
r_n=n^2r_1
$$
Step 3
3 of 4
Now, we can insert the value of the radius of the ground orbit in helium atom and take that $n=2$ so we have that
$$
r_2=4r_1=4times 0.0265times 10^{-9}=boxed{0.106times 10^{-9}textrm{ m}}
$$
Result
4 of 4
$$
r_2=0.106times 10^{-9}textrm{ m}
$$
r_2=0.106times 10^{-9}textrm{ m}
$$
Exercise 13
Step 1
1 of 1
We can obtain the absorption spectrum of the gas by using its sample as a particular prism and let the regular white light through it. Now, on the screen behind the sample the light appears with dark lines which correspond to the wavelengths absorbed by the gas atoms.
Exercise 14
Step 1
1 of 5
In this problem we have a transition between two given levels and we are asked to find the emitted photon wavelength and determine its position in the EM spectrum.
Step 2
2 of 5
In Bohr model, the energy level value is defined by the following formula
$$
E_n=-frac{13.6}{n^2}
$$
So the transition between two levels is
$$
E_{nm}=E_n-E_m=-13.6times(frac{1}{n^2}-frac{1}{m^2})
$$
Step 3
3 of 5
Now, we can take the values given in the problem and insert it into the relation above
$$
E_{101-100}=-13.6times (frac{1}{101^2}-frac{1}{100^2})=2.68times 10^{-5}textrm{ eV}
$$
Step 4
4 of 5
Now we can use the Planck formula to get the wavelength using the fact that $hc=1240times 10^{-9}$eV$cdot$m
$$
lambda=frac{hc}{E}=frac{1240times 10^{-9}}{2.68times 10^{-5}}=boxed{4.6times 10^{-2} textrm{ m}}
$$
Which belongs to microwave range.
Result
5 of 5
$$
lambda=4.6times 10^{-2} textrm{ m; microwave}
$$
lambda=4.6times 10^{-2} textrm{ m; microwave}
$$
Exercise 15
Step 1
1 of 4
In this problem we are asked a hypothetical question of how far away would the electron be of a ball of 5cm radius was used as the nucleus.
Step 2
2 of 4
To solve this problem we are going to use the ratio of the values of different quantities. Let’s $r_0$ is the diameter of the nucleus, $a_0$ is the radius of the hydrogen atom in its ground state we have that
$$
frac{r_0}{a_0}=frac{r_{ball}}{X}
$$
Step 3
3 of 4
From the above expression one can easily express the hypothetical distance to the electron as
$$
X=frac{r_{ball}}{d_0}a_0=frac{5times 10^{-2}}{1.5times 10^{-15}}times 5.3times 10^{-11}
$$
Finally, we have that
$$
boxed{X=1.77times 10^3textrm{m}}
$$
Result
4 of 4
$$
X=1.77times 10^3textrm{m}
$$
X=1.77times 10^3textrm{m}
$$
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