# 11.E: Quantum Mechanics and Atomic Structure (Exercises)

- Page ID
- 41337

### Q11.1

What is the wavelength associated with a photon of a light with the energy is \(3.6 \times 10^{-19}\;J\)?

### S11.1

\[E= hv = \dfrac{hc}{λ} \]

\[3.6 \times 10^{-19}\; J = \dfrac{(6.626 \times 10^{-34}\; J\;s) ( 3 \times 10^8\; ms^{-1})}{\lambda}\]

\[λ = 0.55 \times 10^{-7}\; m = 550\; nm\]

### Q11.2

Calculate the energy of a photon of a light with the frequency is \(6.5 \times 10^{-14}\; s^{-1}\) ?

### S11.2

\[E= hv\]

\[E=( 6.626 \times 10^{-34}J\;s) ( 6.5 \times 10^{-14} s^{-1})\]

\[E= 4.3 \times 10^{-19}\;J\]

### Q11.3a

What are the wavelength of the light emitted by atomic hydrogen when the electron move from \(n=5\) to \(n=3\) orbit in the Bohr model?

### S11.3a

Bohr model:

\[\dfrac{1}{λ} = R_H \left(\dfrac{1}{n_a^2} -\dfrac{1}{n_b^2} \right)\]

\[\dfrac{1}{λ} = ( 1.09678 \times 10^{-2} nm^{-1} ) \left(\dfrac{1}{9}- \dfrac{1}{25}\right) \]

\[λ = 1,282\; nm\]

### Q11.3b

### S11.3b

\[\Delta E=-hcR_H \left(\dfrac{1}{3^2}\ - \dfrac{1}{4^2} \right) \]

\[\Delta E=-6.626 \times 10^{-34} J.s)(2.9979 \times 10^8\; m/s)(109,737\; cm^{-1}) \left(\dfrac{1}{3^2} - \dfrac{1}{4^2}\right))\]

\[\Delta E= 1.8 \times 10^{-21}\; J\]

### Q11.4

Calculate the wavenumber of the wavelength of the light emitted from the \(n=8\) to \(n=6\) transition.

### S11.4

This is a simple application of the Rydberg equation

\[ \tilde{\nu} R_H \left (\dfrac{1}{n_{f}^{2}} -\dfrac{1}{n_{i}^{2}} \right)\]

with \(n_i = 6\) and \(n_f = 8\).

\[ \tilde{\nu}=109,737cm^{-1}\left (\dfrac{1}{6^{2}}-\dfrac{1}{8^{2}} \right)=1333.61 \; cm^{-1}\]

### Q11.5

Calculate the wavelength associated with a 42 g baseball with speed of 80 m/s.

### S11.5

\[ \lambda =\dfrac{h}{p} = \dfrac{h}{mv}\]

\[ \lambda = \dfrac{6.626 \times 10^{34} J s}{(0.042 \; kg)( 80 m/s)} = 1.97 \times 10^{-34} \, m\]

This is a very small wavelength as expected since a ball behaves rather classically (i.e., non-quantum).

### Q11.6

Calculate the energy of a 530-nm photon.

### S11.6

\[p =\dfrac{E}{c}\]

\[\lambda = \dfrac{h}{p}\]

\[\lambda = \dfrac{hc}{E}\]

\[ E= \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 1-^{-34})( 3 \times 10^{8})}{530 \times 10^{-9}} = 3.75 \times 10^{-9} \; J\]

## 11.3: The Photoelectric Effect

### Q11.7

### S11.7

- The number of electrons ejected is proportional to the intensity of light
- The kinetic energy of the ejected electrons is proportional to the frequency of the ejected light
- No electrons can be ejected if the frequency of the light is lower than a certain value, called the threshold frequency (\(nu_o\)).

### Q11.8a

### S11.8a

### Q11.8b

*first*ionization of an atom to be an endothermic process or an exothermic process?

### S11.8b

### Q11.9a

### S11.9a

### Q11.9b

### S11.9b

^{-31}kg

^{9}m/s

### Q11.10a

^{0}. FInd uncertainty in its velocity?

### S11.10a

Δp ≥ h/ 4πΔx = 6.626 x 10^{-34} Js/ 4π(0.5 x 10^{-10}m) = 1.311.05 x 10^{-24}kg m s^{-1}

Δ v= Δp/ m ≥ 1.3 x 10^{-24 }kg m s^{-1}/ 9.109 x 10^{-31 }kg = 1 x 10^{6 }m s^{-1}

**Q11.10**

If the uncertainty of measuring the position of an electron is 2.0 Å, what is the uncertainty of simultaneously measuring its velocity? *Hint: What formula deals with uncertainty of measurements?*

## 11.7: The Schrödinger Wave Equation

### Q11.11

A typical mass for a horse is 510 kg, and a typical galloping speed is 22 kilometers per hour. Use these values to answer the following questions.

- What is the momentum of a galloping horse? What is its wavelength?
- If a galloping horse's velocity and position are simultaneously measured, and the velocity is measured to within ± 1.0%, what is the uncertainty of its position?
- Suppose Planck's constant was actually 0.01 J s. How would that change your answers to (a) and (b)? Which values would be unchanged?

*Hints:*

*de Broglie's postulate deals with the wave-like properties of particles.**Heisenberg's uncertainty principle deals with uncertainty of simultaneous measurements.*

### S11.11

(a)

1. Use the relationship between velocity and momentum to find the momentum.

$$ p = m v $$

$$ p = 510\ kg \times 22\ \dfrac{km}{hr} \times \dfrac{1000\ m}{km} \times \dfrac{hr}{3600\ s} $$

$$ p = 3.1 \times 10^3\ \dfrac{kg\ m}{s} $$

2. Find the de Broglie wavelength.

$$ \lambda = \dfrac{h}{p} $$

$$ \lambda = \dfrac{6.626 \times 10^{-34}\ J\ s}{3.1 \times 10^3\ \dfrac{kg\ m}{s}} \times \dfrac{\dfrac{kg\ m^2}{s^2}}{J} $$

$$ \lambda = 2.1 \times 10^{-37}\ m $$

(b)

1. Find the uncertainty of momentum from the uncertainty of velocity.

$$ \Delta p = m \Delta v $$

$$ \Delta v = 0.01 \times v $$

$$ \Delta p = 510\ kg \times 0.01 \times 22\ \dfrac{km}{hr} \times \dfrac{1000\ m}{km} \times \dfrac{hr}{3600\ s} $$

$$ \Delta p = 31 \dfrac{kg\ m}{s} $$

2. Use Heisenberg's uncertainty principle to find the uncertainty of position.

$$ \Delta x \Delta p \geq \dfrac{h}{4 \pi} $$

$$ \Delta x \geq \dfrac{h}{4 \pi \Delta p} $$

$$ \Delta x \geq \dfrac{6.626 \times 10^{-34}\ J\ s}{4 \pi \times 3.1 \times 10^3 \dfrac{kg\ m}{s}} \times \dfrac{\dfrac{kg\ m^2}{s^2}}{J} $$

$$ \Delta x \geq 1.7 \times 10^{-36}\ m $$

(c)

The calculations that involve Planck's constant (h) will change. This much larger value will make the wave-like properties of the horse more important; its wavelength and uncertainty of position will increase dramatically:

$$ \lambda ' = \dfrac{h'}{p} = \dfrac{0.010\ J\ s}{3.1 \times 10^3\ \dfrac{kg\ m}{s} \times \dfrac{\dfrac{kg\ m^2}{s^2}}{J}} $$

$$ \lambda ' = 3.2 \times 10^{-6}\ m = 3200\ nm $$

$$ \Delta x' \geq \dfrac{h'}{4 \pi \Delta p} = \dfrac{0.010\ J\ s}{4 \pi \times 3.1 \times 10^3 \dfrac{kg\ m}{s}} \times \dfrac{\dfrac{kg\ m^2}{s^2}}{J} $$

$$ \Delta x' \geq 2.6 \times 10^{-5} m = 26\ \mu m $$

The momentum would not change.

### Q11.12a

### S11.12a

p= h/ λ = 6.626 x 10 ^{-34 }J s/ 0.5

= 3.313 x 10^{-34} kg m s^{-1}

V= p/m = 3.313 x 10^{-34 }/ 42 kg = 7.88 x 10^{-36 }ms^{-1}

### Q11.12b

When a particle passes through a slit, diffraction occurs if the particle's wavelength is on the same order of magnitude as the width of the slit. At approximately what velocity would a 1670 kg car have to move through the Lincoln Tunnel (6.6 m width) for diffraction to occur? *Hint: **How can you find the wavelength of the car?*

### S11.12b

1. Use de Broglie's equation to relate velocity to wavelength; solve for velocity.

$$ \lambda = \dfrac{h}{m v} $$

$$ v = \dfrac{h}{m \lambda} $$

2. Set the wavelength equal to the tunnel width and solve.

$$ v = \dfrac{6.626 \times 10^{-34}\ J\ s}{1670\ kg \times 6.6\ m} \times \dfrac{\dfrac{kg\ m^2}{s^2}}{J} $$

Answer: the car would have to be driving *very* slowly.

$$ v = 6.0 \times 10^{-38}\ m/s $$

### Q11.13a

The Paschen emission spectrum is the collection of spectral lines emitted by H atoms, where the final state is n = 3. What are the shortest and longest wavelengths (in nm) of Paschen spectral lines? *Hints: T he minimum value for n_{i} is 4, the maximum value for n_{i} is infinity and the Rydberg formula describes the emission spectra of hydrogen*

### Q11.13b

### S11.13b

A photon has a specific energy according to the energy between the excited and ground state.

E = (1/nf2 - 1/ni2), A = hv

The lyman series is when nf is the ground state or nf =1 and Balmer series is when nf =2

nf = 1 the value , ni from 2 to infinity with the values between 1 and ½. For the balmer series ni is from 3 to infinity the range between 1/4 and 5/36. Thus, these intervals don not overlap --> the energies do not overlap --> the lines in the series cant overlap

### Q11.14

^{+}is 9.72x10

^{-18}nm, calculate the wavelength of He

^{+}ions from n=3 to n=2.

### S11.14

\[\dfrac{1}{\lambda}=R_{H} \left | \dfrac{1}{n^{2}_{i}}-\dfrac{1}{n^{2}_{f}} \right | \]

\[\dfrac{1}{\lambda}=9.72\times10^{-18} \left | \dfrac{1}{3^{2}}-\dfrac{1}{2^{2}} \right | \]

\[\lambda=7.41\times10^{17}nm\]

### Q11.15

Derive from the following equation to solve for wavelength.

\(\Delta E=hcR_{H}(\dfrac{1}{n_{i}^{2}}-\dfrac{1}{n_{f}^{2}})\)

### S11.15

\[\Delta E=R_{H} \left | \dfrac{1}{n^{2}_{i}}-\dfrac{1}{n^{2}_{f}} \right | \]

\[\dfrac{v}{c}=\Delta E=R_{H} \left | \dfrac{1}{n^{2}_{i}}-\dfrac{1}{n^{2}_{f}} \right | \]

\[\dfrac{1}{\lambda}=\dfrac{v}{c}=\Delta E=R_{H} \left | \dfrac{1}{n^{2}_{i}}-\dfrac{1}{n^{2}_{f}} \right | \]

### Q11.16

^{20 }numbers of photons, how much energy in joules was that one flash?

### Q11.50

### S11.50

### Q11.51

A scientist determined the kinetic energy released electrons of cesium metal through a photoelectric experiment. Determine h and the \work function for cesium graphically from the following results:

λ/nm |
410 | 420 | 475 | 500 | 550 | 630 |

V/volt |
1.6 | 1.4 | 1.1 | 0.91 | 0.72 |
0.37 |

### S11.51

First, convert wavelength to frequency and volt to energy in J.

f (1/sec) |
Energy (J) |

\(7.31707 \times 10^{15}\) | 1.6 |

\(7.14286 \times 10^{15}\) | 1.4 |

\(6.3157 \times 10^{15}\) | 1.1 |

\(6 \times 10^{15}\) | 0.91 |

\(5.45455 \times 10^{15}\) | 0.72 |

\(4.7619 \times 10^{15}\) | 0.32 |

Then graph the values and determine the linear equation.

The linear equation is

\[y=4.7 \times 10^{-16} f - 1.89\]

Use the equation below:

hf-=E_{k} (max kinetic energy)

h=slope

= y-intercept

h= 4.7 X 10^{-16}

= 1.89

### Q11.52

Calculate the de Broglie wavelength of a Cl_{2} molecule at 300 K.

### S11.52

\[ v_{rms} = \sqrt{\dfrac{3RT}{M}}\]

=((3)(8.314)(300)/(70.8X10^{-3} kg) = 325 m/s

\[ \lambda = \dfrac{h}{mv_{rms}}\]

=(6.626X10^{-34 }J*s)/[(70.8 amu)(1.66X10^{-27} kg/amu)(325 m/s)] = 1.73 X 10^{-11} m

### Q11.53

Consider a balloon with a diameter of \(2.5 \times 10^{-5}\; m\). What is the uncertainty of the velocity of an oxygen molecule that is trapped inside.

### S11.53

Use the direct application of the uncertainty principle:

\[ \Delta x \Delta p \ge \dfrac{h}{4 \pi}\]

Let's consider the molecule has an uncertainly that is \(\pm\, radius\) of the balloon.

\[\Delta x = 1.3 \times 10^{-5}\; m\]

The uncertainty of the momentum of the molecule can be estimated via the uncertainty principle.

\[ \Delta p =\dfrac{h}{4 \pi \Delta x} = \dfrac{(6.626 \times 10^{-34} J*s)}{(4\pi) (1.3 \times 10^{-5}\; m)} = 4.05 \times 10^{-30} kg\,m\,s^{-1}\]

This can be converted to uncertainty in velocity via

\[p=mv\]

or

\[ \Delta p =m_e \Delta v\]

with the electron mass \(m_e\) equal to \(9.109 \times 10^{-31}\; kg\). since the mass of the molecule is **not **uncertain.

\[ \Delta v = \dfrac{ \Delta p }{ m_e } = \dfrac{4.05 \times 10^{-30}}{9.109 \times 10^{-31}\; kg} = 4.5\; m/s\]