Home Page Textbooks Physics: Principles and Problems Page 773: Standardized Test Practice

Physics: Principles and Problems

9th Edition

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

ISBN: 9780078458132

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Page 773: Standardized Test Practice

Exercise 1

Step 1

1 of 2

The nuclear model, also known as the planetary model was based on the Rutherford’s experiment. Therefore the correct answer is B.

Result

2 of 2

The correct answer is B.

Exercise 2

Step 1

1 of 3

In this problem, we are given an emission from the mercury atom of a known wavelength for which we should find the corresponding energy. We can do so by using Planck’s formula and inserting $hc=1240times 10^{-9}$ eV$cdot$m.

$$
E=frac{hc}{lambda}
$$

Step 2

2 of 3

center{After we plug in the values we have that }
[E=frac{1240times 10^{-9}}{405times 10^{-9}}=3.06textrm{ eV}]
so the answer is C. \

Result

3 of 3

The correct answer is C.

Exercise 3

Step 1

1 of 4

In this problem we are given the mercury atom that undergoes the deexcitation process from $E_7$to $E_4$ while emitting a photon. We should find the wavelength of the emitted photon.

Step 2

2 of 4

We can do so by using the Planck equation to express the wavelength as follows

$$
lambda=frac{hc}{E_{ph}}
$$

where $hc=1240times 10^{-9}$eV$cdot$m. However, we should find the energy first and we can do that by looking at the diagram of the mercury energy states.

$$
E_{ph}=abs{E_4-E_7}=abs{-4.95+2.48}=2.47textrm{ eV}
$$

Step 3

3 of 4

Now, one can simply insert the given values into the expression for the wavelength to have that

$$
lambda=frac{1240times 10^{-9}}{2.47}=boxed{502textrm{ nm}}
$$

so the correct answer is D.

Result

4 of 4

$$
textrm{D) }lambda=502textrm{ nm}
$$

Exercise 4

Solution 1

Solution 2

Step 1

1 of 2

B) The locations of the electrons around the nucleus are known precisely.

Result

2 of 2

Step 1

1 of 1

$mathrm{B}$ is the wrong statement. This is due to the axioms of quantum mechanics. Mainly, we don’t represent a state of a system with precise coordinates (Positions of electrons in this example) and their change in time like we do in classical mechanics.

In quantum mechanics we say that a state of system is represented by a state function $Psi$, which is a function of time and coordinates. Due to Heisenberg’s uncertainty principle, in quantum mechanics, we don’t talk about precise coordinates, but the probability of finding a object in a certain place.

Exercise 5

Step 1

1 of 2

Since the energy of a photon is directly proportional to its frequency according to Planck’s formula we have that the highest energy transition will lead to the highest frequency photon emission. From the graph, we see that $E_{6rightarrow2}$ transition has the largest energy so it will lead to the highest frequency photons. So the correct answer is D.

Result

2 of 2

The correct answer is D.

Exercise 6

Step 1

1 of 4

In this problem, we are given the hydrogen atom that undergoes the deexcitation process from $E_4$to $E_2$ while emitting a photon. We should find the frequency of the emitted photon.

Step 2

2 of 4

We can do so by using the Planck equation to express the frequency as follows

$$
f=frac{E_{ph}}{h}
$$

However, we should find the energy first and we can do that by looking at the diagram of the mercury energy states.

$$
E_{ph}=abs{E_2-E_4}=abs{-3.4+0.85}=2.55textrm{ eV}
$$

Step 3

3 of 4

Now, one can simply insert the given values into the expression for the frequency while converting the energy to Joules to have that

$$
f=frac{2.55 times 1.6times 10^{-19}}{6.63times 10^{-34}}=boxed{6.15times 10^{14}textrm{ Hz}}
$$

so the correct answer is C.

Result

4 of 4

$$
textrm{C) }f=6.15times 10^{14}textrm{ Hz}
$$

Exercise 7

Step 1

1 of 4

In this problem, we are given the hydrogen atom that undergoes the deexcitation process from $E_5$to $E_2$ while emitting a photon. We should find the wavelength of the emitted photon.

Step 2

2 of 4

We can do so by using the Planck equation to express the wavelength as follows

$$
lambda=frac{hc}{E_{ph}}
$$

where $hc=1240times 10^{-9}$eV$cdot$m. However, we should find the energy first and we can do that by looking at the diagram of the mercury energy states.

$$
E_{ph}=abs{E_2-E_5}=abs{-3.40+0.54}=2.86textrm{ eV}
$$

Step 3

3 of 4

Now, one can simply insert the given values into the expression for the wavelength to have that

$$
lambda=frac{1240times 10^{-9}}{2.86}=boxed{434textrm{ nm}}
$$

Result

4 of 4

$$
lambda=434textrm{ nm}
$$

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