Content check — College Physics, Chapter 30: Atomic Physics

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Chapter 30: Atomic Physics

38 problems · 5 figures

Using the given charge-to-mass ratios for electrons and protons, and knowing the magnitudes of their charges are equal, what is the ratio of the proton’s mass to the electron’s? (Note that since the charge-to-mass ratios are given to only three-digit accuracy, your answer may differ from the accepted ratio in the fourth digit.)

Answer

1 . 84 \times 10^{3}

30-3 §30.1 source ↗

If someone wanted to build a scale model of the atom with a nucleus 1.00 m in diameter, how far away would the nearest electron need to be?

Answer

50 km

30-4 §30.2 source ↗

Rutherford found the size of the nucleus to be about

10^{- 15} m

. This implied a huge density. What would this density be for gold?

Answer

6 \times 10^{20} {kg/m}^{3}

30-6 §30.2 source ↗

(a) An aspiring physicist wants to build a scale model of a hydrogen atom for her science fair project. If the atom is 1.00 m in diameter, how big should she try to make the nucleus?

(b) How easy will this be to do?

Answer

(a) $10.0 μm$

(b) It isn’t hard to make one of approximately this size. It would be harder to make it exactly $10.0 μm$ .

30-7 §30.3 source ↗

By calculating its wavelength, show that the first line in the Lyman series is UV radiation.

Answer

$\frac{1}{λ} =$ $R \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \Rightarrow λ =$ $\frac{1}{R} \frac{(n_{i} \cdot n_{f})^{2}}{n_{i}^{2} - n_{f}^{2}}; n_{i} =$ $2, n_{f} =$ $1,$ so that

$λ =$ $\frac{m}{1.097 \times 10^{7}} \frac{(2 \times 1)^{2}}{2^{2} - 1^{2}} =$ $1.22 \times 10^{- 7} m =$ $122 nm$ , which is UV radiation.

30-9 §30.3 source ↗

Look up the values of the quantities in

a_{B} =

\frac{h^{2}}{{4 π}^{2} m_{e} {kq}_{e}^{2}}

, and verify that the Bohr radius

a_{B}

0.529 \times 10^{- 10} m

Answer

a_{B} =

\frac{h^{2}}{{4 π}^{2} m_{e} {kZq}_{e}^{2}} =

\frac{(6.626 \times 10^{- 34} J·s)^{2}}{{4 π}^{2} (9.109 \times 10^{- 31} kg) (8.988 \times 10^{9} N · m^{2} / C^{2}) (1) (1.602 \times 10^{- 19} C)^{2}} =

0.529 \times 10^{- 10} m

30-11 §30.3 source ↗

If a hydrogen atom has its electron in the

n = 4

state, how much energy in eV is needed to ionize it?

Answer

0.850 eV

30-13 §30.3 source ↗

Find the radius of a hydrogen atom in the

n = 2

state according to Bohr’s theory.

Answer

2.12 \times 10^{–10} m

30-15 §30.3 source ↗

What is the smallest-wavelength line in the Balmer series? Is it in the visible part of the spectrum?

Answer

365 nm

It is in the ultraviolet.

30-17 §30.3 source ↗

Do the Balmer and Lyman series overlap? To answer this, calculate the shortest-wavelength Balmer line and the longest-wavelength Lyman line.

Answer

No overlap

365 nm

122 nm

30-19 §30.3 source ↗

A wavelength of

4 . 653 μm

is observed in a hydrogen spectrum for a transition that ends in the

n_{f} = 5

level. What was

n_{i}

for the initial level of the electron?

Answer

30-21 §30.3 source ↗

A beryllium ion with a single electron (denoted ${Be}^{3 +}$ ) is in an excited state with radius the same as that of the ground state of hydrogen.

(a) What is $n$ for the ${Be}^{3 +}$ ion?

(b) How much energy in eV is needed to ionize the ion from this excited state?

Answer

(a) 2

(b) 54.4 eV

30-23 §30.3 source ↗

Verify Equations

r_{n} = \frac{n^{2}}{Z} a_{B}

and

a_{B} =

\frac{h^{2}}{{4 π}^{2} m_{e} {kq}_{e}^{2}} =

0.529 \times 10^{- 10} m

using the approach stated in the text. That is, equate the Coulomb and centripetal forces and then insert an expression for velocity from the condition for angular momentum quantization.

Answer

\frac{{kZq}_{e}^{2}}{r_{n}^{2}} =

\frac{m_{e} V^{2}}{r_{n}},

so that

r_{n} =

\frac{{kZq}_{e}^{2}}{m_{e} V^{2}} =

\frac{{kZq}_{e}^{2}}{m_{e}} \frac{1}{V^{2}} .

From the equation

m_{e} {vr}_{n} = n \frac{h}{2 π},

we can substitute for the velocity, giving:

r_{n} =

\frac{{kZq}_{e}^{2}}{m_{e}} \cdot \frac{{4 π}^{2} m_{e}^{2} r_{n}^{2}}{n^{2} h^{2}}

so that

r_{n} =

\frac{n^{2}}{Z} \frac{h^{2}}{{4 π}^{2} m_{e} {kq}_{e}^{2}} =

\frac{n^{2}}{Z} a_{B},

where

a_{B} =

\frac{h^{2}}{{4π}^{2} m_{e} {kq}_{e}^{2}}

30-25 §30.4 source ↗

(a) What is the shortest-wavelength x-ray radiation that can be generated in an x-ray tube with an applied voltage of 50.0 kV? (b) Calculate the photon energy in eV. (c) Explain the relationship of the photon energy to the applied voltage.

Answer

(a) $0.248 \times 10^{−10} m$

(b) 50.0 keV

(c) The photon energy is simply the applied voltage times the electron charge, so the value of the voltage in volts is the same as the value of the energy in electron volts.

30-27 §30.4 source ↗

An x ray tube has an applied voltage of 100 kV. (a) What is the most energetic x-ray photon it can produce? Express your answer in electron volts and joules. (b) Find the wavelength of such an X–ray.

Answer

(a) $100 \times 10^{3} eV$ , $1 . 60 \times 10^{- 14} J$

(b) $0.124 \times 10^{- 10} m$

30-29 §30.4 source ↗

What are the approximate energies of the

K_{α}

and

K_{β}

x rays for copper?

Answer

(a) 8.00 keV

(b) 9.48 keV

30-30 §30.5 source ↗

Figure 30.37 shows the energy-level diagram for neon. (a) Verify that the energy of the photon emitted when neon goes from its metastable state to the one immediately below is equal to 1.96 eV. (b) Show that the wavelength of this radiation is 633 nm. (c) What wavelength is emitted when the neon makes a direct transition to its ground state?

On the left side of the figure, the ground state and first metastable state of helium atom are shown, and on the right side, the ground state and first metastable state of neon atom are shown. The difference between the two states of helium and neon atoms are estimated to be twenty point six one electron volts and twenty point six six electron volts, respectively. The collision transfer energy from helium to neon atoms is given as one point nine six electron volts. — Figure 30.37 in the book ↗

Answer

(a) 1.96 eV

(b) $(1240 eV·nm) / (1 . 96 eV) = 633 nm$

30-32 §30.5 source ↗

Ruby lasers have chromium atoms doped in an aluminum oxide crystal. The energy level diagram for chromium in a ruby is shown in Figure 30.59. What wavelength is emitted by a ruby laser?

The figure shows energy levels of chromium atoms in an aluminum oxide crystal. Ground state is at zero point zero electron volts, first metastable state is at one point seventy nine electron volts, second state is at two point three electron volts, and the third state is at three point zero electron volts. — Figure 30.59 in the book ↗

Answer

693 nm

30-34 §30.5 source ↗

Some of the most powerful lasers are based on the energy levels of neodymium in solids, such as glass, as shown in Figure 30.60. (a) What average wavelength light can pump the neodymium into the levels above its metastable state? (b) Verify that the 1.17 eV transition produces

1 . 06 μm

radiation.

The figure shows different energy levels of neodymium atoms in glass. The ground state is at zero electron volts, first state is at zero point five zero electron volts, the metastable second state is at one point sixty seven electron volts, and the group state levels above metastable second are at two point one electron volts. The photons release one point seventeen electron volts at wavelength of one point zero six micro meters while coming from the metastable second state to first state. — Figure 30.60 in the book ↗

Answer

(a) 590 nm

(b) $(1240 eV·nm) / (1 . 17 eV) = 1.06 μm$

30-35 §30.8 source ↗

If an atom has an electron in the

n = 5

state with

m_{l} = 3

, what are the possible values of

l

Answer

l = 4, 3

are possible since

l < n

and

∣ m_{l} ∣ ≤ l

30-37 §30.8 source ↗

What are the possible values of

m_{l}

for an electron in the

n = 4

state?

Answer

n =

4 \Rightarrow l =

3, 2, 1, 0 \Rightarrow m_{l} =

\pm 3, \pm 2, \pm 1, 0

are possible.

30-39 §30.8 source ↗

(a) Calculate the magnitude of the angular momentum for an

l = 1

electron. (b) Compare your answer to the value Bohr proposed for the

n = 1

state.

Answer

(a) $1 . 49 \times 10^{- 34} J \cdot s$

(b) $1 . 06 \times 10^{- 34} J \cdot s$

30-41 §30.8 source ↗

Repeat Exercise 30.40 for

l = 3

Answer

(a) $3 . 66 \times 10^{- 34} J \cdot s$

(b) $s =$ $9.13 \times 10^{- 35} J \cdot s$

30-43 §30.8 source ↗

What angles can the spin

s

of an electron make with the

z

-axis?

Answer

θ =

54.7º, 125.3º

30-44 §30.9 source ↗

(a) How many electrons can be in the $n = 4$ shell?

(b) What are its subshells, and how many electrons can be in each?

Answer

(a) 32. (b)

2 in s, 6 in p, 10 in d,

and 14 in

f

, for a total of 32.

30-46 §30.9 source ↗

(a) If one subshell of an atom has 9 electrons in it, what is the minimum value of

l

? (b) What is the spectroscopic notation for this atom, if this subshell is part of the

n = 3

shell?

Answer

(a) 2

(b) $3 d^{9}$

30-48 §30.9 source ↗

Which of the following spectroscopic notations are not allowed? (a)

5 s^{1}

(b)

1 d^{1}

(c)

4 s^{3}

(d)

3 p^{7}

(e)

5 g^{15}

. State which rule is violated for each that is not allowed.

Answer

(b) $n \geq l$ is violated,

(d) cannot have 7 electrons in $p$ subshell since $7 >$ $(2 l + 1) =$ $2 (2 + 1) =$ $6$

30-50 §30.9 source ↗

(a) Using the Pauli exclusion principle and the rules relating the allowed values of the quantum numbers $n, l, m_{l}, m_{s}$ , prove that the maximum number of electrons in a subshell is $2 n^{2}$ .

(b) In a similar manner, prove that the maximum number of electrons in a shell is 2n².

Answer

(a) The number of different values of $m_{l}$ is $\pm l, \pm (l - 1),$ $..., 0$ for each $l > 0$ and one for $l = 0 \Rightarrow (2 l + 1) .$ Also an overall factor of 2 since each $m_{l}$ can have $m_{s}$ equal to either $+ 1 / 2$ or $-$ $1 / 2 \Rightarrow 2 (2 l + 1)$ .

(b) for each value of $l$ , you get $2 (2 l + 1)$

$=$ $0, 1, 2, ..., (n –1) \Rightarrow 2 (2) (0) + 1 + (2) (1) + 1 + . . . . + (2) (n - 1) + 1 =$ $\underset{\underset{n terms}{︸}}{2 1 + 3 + . . . + (2 n - 3) + (2 n - 1)}$ to see that the expression in the box is $= n^{2},$ imagine taking $(n - 1)$ from the last term and adding it to first term $=$ $2 1 + (n –1) + 3 + . . . + (2 n - 3) + (2 n - 1) - (n - 1) =$ $2 n + 3 + . . . . + (2 n - 3) + n .$ Now take $(n - 3)$ from penultimate term and add to the second term $2 \underset{\underset{n terms}{︸}}{n + n + . . . + n + n} =$ ${2 n}^{2}$ .

30-52 §30.9 source ↗

Integrated Concepts

The electric and magnetic forces on an electron in the CRT in Figure 30.7 are supposed to be in opposite directions. Verify this by determining the direction of each force for the situation shown. Explain how you obtain the directions (that is, identify the rules used).

Image of a cathode ray tube on x axis between two inverted L shaped north and south pole magnets on y axis, with z axis as a wire carrying high voltage supply to the charging plates inside the C R T. Zoomed image of the charging plate area inside the C R T showing the intersection of magnetic field between the poles in red lines towards south pole on the y axis along with an electron beam in green color line with velocity v toward right on the x axis. — Figure 30.7 in the book ↗

Answer

The electric force on the electron is up (toward the positively charged plate). The magnetic force is down (by the RHR).

30-54 §30.9 source ↗

Integrated Concepts

A galaxy moving away from the earth has a speed of $0.0100 c$ . What wavelength do we observe for an $n_{i} = 7$ to $n_{f} = 2$ transition for hydrogen in that galaxy?

Answer

401 nm

30-56 §30.9 source ↗

Integrated Concepts

In a Millikan oil-drop experiment using a setup like that in Figure 30.9, a 500-V potential difference is applied to plates separated by 2.50 cm. (a) What is the mass of an oil drop having two extra electrons that is suspended motionless by the field between the plates? (b) What is the diameter of the drop, assuming it is a sphere with the density of olive oil?

Image of the apparatus used in the Millikan oil drop experiment, consisting of a parallel pair of horizontal metal plates with a pin hole opening in the top plate. The top plate has positive charge and the bottom plate has negative charge. Picture of a flashlight as a bright source of light and a beam of light passing in between the plates from left is shown. A telescope is shown at the front and an oil atomizer above the positive plate is also depicted. A zoomed image of metal plates describing the force acting on the oil droplet is also shown. Arrows pointing upwards are forces of electric field while arrows pointing downwards depict the force of gravity. — Figure 30.9 in the book ↗

Answer

(a) $6 . 54 \times 10^{- 16} kg$

(b) $5 . 54 \times 10^{- 7} m$

30-58 §30.9 source ↗

Integrated Concepts

In a laboratory experiment designed to duplicate Thomson’s determination of $q_{e} / m_{e}$ , a beam of electrons having a velocity of $6 . 00 \times 10^{7} m/s$ enters a $5.00 \times 10^{- 3} T$ magnetic field. The beam moves perpendicular to the field in a path having a 6.80-cm radius of curvature. Determine $q_{e} / m_{e}$ from these observations, and compare the result with the known value.

Answer

1 . 76 \times 10^{11} C/kg

, which agrees with the known value of

1 . 759 \times 10^{11} C/kg

to within the precision of the measurement

30-60 §30.9 source ↗

Integrated Concepts

Particles called muons exist in cosmic rays and can be created in particle accelerators. Muons are very similar to electrons, having the same charge and spin, but they have a mass 207 times greater. When muons are captured by an atom, they orbit just like an electron but with a smaller radius, since the mass in $a_{B} =$ $\frac{h^{2}}{{4π}^{2} m_{e} {kq}_{e}^{2}} =$ $0.529 \times 10^{- 10} m$ is 207 $m_{e}$ .

(a) Calculate the radius of the $n = 1$ orbit for a muon in a uranium ion ( $Z = 92$ ).

(b) Compare this with the 7.5-fm radius of a uranium nucleus. Note that since the muon orbits inside the electron, it falls into a hydrogen-like orbit. Since your answer is less than the radius of the nucleus, you can see that the photons emitted as the muon falls into its lowest orbit can give information about the nucleus.

Answer

(a) 2.78 fm

(b) 0.37 of the nuclear radius.

30-62 §30.9 source ↗

Integrated Concepts

A carbon dioxide laser used in surgery emits infrared radiation with a wavelength of $10.6 μm$ . In 1.00 ms, this laser raised the temperature of ${1.00 cm}^{3}$ of flesh to $100ºC$ and evaporated it.

(a) How many photons were required? You may assume flesh has the same heat of vaporization as water. (b) What was the minimum power output during the flash?

Answer

(a) $1 . 34 \times 10^{23}$

(b) 2.52 MW

30-64 §30.9 source ↗

Integrated Concepts

(a) An excimer laser used for vision correction emits 193-nm UV. Calculate the photon energy in eV.

(b) These photons are used to evaporate corneal tissue, which is very similar to water in its properties. Calculate the amount of energy needed per molecule of water to make the phase change from liquid to gas. That is, divide the heat of vaporization in kJ/kg by the number of water molecules in a kilogram.

Answer

(a) 6.42 eV

(b) $7.27 \times 10^{- 20} J/molecule$

(c) 0.454 eV, 14.1 times less than a single UV photon. Therefore, each photon will evaporate approximately 14 molecules of tissue. This gives the surgeon a rather precise method of removing corneal tissue from the surface of the eye.

30-66 §30.9 source ↗

Integrated Concepts

A pulsar is a rapidly spinning remnant of a supernova. It rotates on its axis, sweeping hydrogen along with it so that hydrogen on one side moves toward us as fast as 50.0 km/s, while that on the other side moves away as fast as 50.0 km/s. This means that the EM radiation we receive will be Doppler shifted over a range of $\pm 50 .0 km/s$ . What range of wavelengths will we observe for the 91.20-nm line in the Lyman series of hydrogen? (Such line broadening is observed and actually provides part of the evidence for rapid rotation.)

Answer

91.18 nm to 91.22 nm

30-68 §30.9 source ↗

Unreasonable Results

(a) What voltage must be applied to an X-ray tube to obtain 0.0100-fm-wavelength X-rays for use in exploring the details of nuclei? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

Answer

(a) $1 . 24 \times 10^{11} V$

(b) The voltage is extremely large compared with any practical value.

30-72 §30.9 source ↗

Critical Thinking (a) For carbon, calculate the energy when an electron falls from

n = 3

n = 2

, then from

n = 2

n = 1

. Then add these for the total energy released in the process. (b) Calculate the energy released when the electron falls directly from

n = 3

n = 1

. (c) Are the energies in part a and in part b the same?

Answer

(a) $\begin{array}{l} E = - \frac{z^{2}}{n^{2}} \\ E_{3} - E_{2} = - \frac{6^{2}}{3^{2}} - \frac{6^{2}}{2^{2}} 13.6 e V = 68.0 e V \\ E_{3} - E_{2} = - \frac{6^{2}}{2^{2}} - \frac{6^{2}}{1^{2}} 13.6 e V = 367 e V \\ E = 435 e V \end{array}$

(b) $E_{3} -$ $E_{2} =$ $-$ $\frac{6^{2}}{3^{2}} - \frac{6^{2}}{1^{2}} 13.6 e V =$ $435 e V$

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Problems and answers are adapted from College Physics 2e by Paul Peter Urone, Roger Hinrichs, Kim Dirks and Manjula Sharma, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.