# Exercises Solutions


## 4.1: The Wave Theory of Light

### Q4.1.1

$E= hv = \dfrac{hc}{λ} \nonumber$

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

$λ = 0.55 \times 10^{-7}\; m = 550\; nm\nonumber$

### Q4.1.2

$E= hv\nonumber$

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

$E= 4.3 \times 10^{-19}\;J\nonumber$

### Q4.1.3

a. This is a two part problem. The first part is to calculate the energy per photon. The second part of to calculate the flux (number of photon/ time) from the power (energy/time).

• Energy per photon

$E={hv}$ $E=\dfrac{hc} {λ}$

$E=\dfrac{(6.634\times 10^{-34} J·s)·(3\times 10^8 m/s)} {550\times 10^9 m}$

$E=3.618\times 10^{-37} J$

• Number of photons emitted per minute

$\dfrac{100} {E}\ ·\dfrac{60s}{1 min}$

$\dfrac{100} {3.618\times 10^{-37} J}·\dfrac{60s}{1min}$

$=1.658\times 10^{40} \dfrac{photons}{min}$

b. This is a two part problem. The first part is to calculate the energy per photon. The second part of to calculate the flux (number of photon/ time) from the power (energy/time).

• Energy per photon

$E=\dfrac{(6.634\times 10^{-34} J·s)·(3\times 10^8 m/s)} {0.01 m}$

$E=1.99\times 10^{37}J$

• Number of photons emitted per minute

$\dfrac{1000} {1.99\times 10^{37} J}·\dfrac{60s}{1min}$

$=3.02\times 10^{-33} \dfrac{photons}{min}$

## 4.3: The Photoelectric Effect

### Q4.3.1

Since $$\nu=(c/\lambda)$$ Let's express the formula in the following form:

$(1/2)m_ev^2=(hc)/\lambda-\phi$

Where:

• m_e = mass of the electron
• h = Plank's constant

Then:

$v^2=(2/m_e)(hc/\lambda)-\phi$

For 25eV:

$$25eV=4.01\times 10^{-18} J$$

We now plug values inside the formula to calculate velocity:

$v^2=\dfrac{2}{9.11\times 10^{-31} Kg}(\dfrac{6.63 \times10^{-34} J*s \times 3\times10 ^8 m/s}{1.25\times 10^{-9}m}-4.01\times 10^{-18}J)$

From there we get the velocity:

$v=1.84 \times 10^7 m/s$

For 125eV :

$$125eV=2.00\times 10^{-17} J$$

We plug values inside the formula like before:

$v^2=\dfrac{2}{9.11\times 10^{-31} Kg}(\dfrac{6.63 \times10^{-34} J*s \times 3\times10 ^8 m/s}{1.25\times 10^{-9}m}-2.00\times 10^{-17} J)$

From there we obtain velocity:

$v=1.75\times 10^7 m/s$

For 450eV :

$$425eV=6.81\times 10^{-17} J$$

We plug for this values again:

$v^2=\dfrac{2}{9.11\times 10^{-31} Kg}(\dfrac{6.63 \times10^{-34} J*s \times 3\times10 ^8 m/s}{1.25\times 10^{-9}m}-6.81\times 10^{-17} J)$

We obtain velocity:

$v=1.42\times 10^7 m/s$

## 4.4: Bohr's Theory of the Hydrogen Emission Spectrum

### Q4.4.2

This is a simple application of the Rydberg equation

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

with $$n_i = 6$$ and $$n_f = 8$$.

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

## 4.5: de Broglie's Postulate

### Q4.5.1

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

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

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

### Q4.5.2

De Broglie wavelength:

$\lambda = \dfrac{h}{P} = \dfrac{h}{mv}$

where P = momentum, h = Plank's constant (6.626*10-34Js), v = velocity, m = mass

a)

$\lambda = \dfrac{6.626 \times 10^{-34} \text{Js}}{(.8 \times 10^{-3} \text{Kg})(340 \text{m/s})} = 2.44 \times 10^{-33} \text{m}$

b)

$\lambda = \dfrac{6.626 \times 10^{-34} \text{Js}}{(10^{-5} \times 10^{-3} \text{Kg})(10^{-5} \text{m/s})} = 6.626 \times 10^{-21} \text{m}$

c)

$\lambda = \dfrac{6.626 \times 10^{-34} \text{Js}}{(10^{-8} \times 10^{-3} \text{Kg})(10^{-8} \text{m/s})} = 6.626 \times 10^{-15} \text{m}$

b)

$\lambda = \dfrac{6.626 \times 10^{-34} \text{Js}}{(9.1 \times 10^{-31} \text{Kg})(4.8^{6} \text{m/s})} = 1.52 \times 10^{-10} \text{m}$

Mass of an electron is 9.1*10-31 kg.

### Q4.5.3

The de Broglie wavelength is determined by

$\lambda = {\dfrac{h}{p}} = {\dfrac{h}{m \upsilon}}\label{1}$

where $$p$$ is momentum, $$m$$ is mass, and $$\upsilon$$ is velocity.

A neutron's velocity is determined by

$\upsilon = (\dfrac{3 k_B T}{m})^{1/2} \label{2}$

Substituting equation $$(2)$$ into $$(1)$$ we get

$\lambda = \dfrac{h}{(3 m k_B T)^{1/2}}\label{3}$

where $$h$$ is Planck's constant and $$k_B$$ is the Boltzmann constant.

We can now solve for our de Broglie wavelength by plugging in $$T=350 K$$ and $$m=1.67 \times 10^{-27} kg$$ (mass of a neutron)

$\lambda = \dfrac{6.626 \times 10^{-34} m^2 kg/s}{(3 (1.67 \times 10^{-27} kg) (1.38 \times 10^{-23} J/K) (350 K))^{1/2}} = 1.455 \times 10^{-10} m$

## 4.6: The Heisenberg Uncertainty Principle

### Q4.6.2

(a)

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

$p = m v \nonumber$

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

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

2. Find the de Broglie wavelength.

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

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

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

(b)

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

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

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

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

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

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

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

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

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

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

(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:

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

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

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

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

The momentum would not change.

### Q4.6.3

Use the direct application of the uncertainty principle:

$\Delta x \Delta p \ge \dfrac{h}{4 \pi}\nonumber$

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

$\Delta x = 1.3 \times 10^{-5}\; m\nonumber$

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

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

This can be converted to uncertainty in velocity via

$p=mv\nonumber$

or

$\Delta p =m_e \Delta v\nonumber$

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

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

## 4.7: The Schrödinger Wave Equation

### Q4.7.1

a)

1. Apply Operator: $\dfrac{d(3e^{-cx^3})}{dx}$Take derivative of the whole function by multiplying the exponential by the derivative of its superscript: $-9cx^{2}e^{-cx^{3}}$
2. Apply Operator: $\dfrac{d^2(3e^{-cx^3})}{dx^2}$Which is the same as: $\dfrac{d(\dfrac{d(3e^{-cx^3})}{dx})}{dx}$Or: $\dfrac{d(-9cx^{2}e^{-cx^{3}})}{dx}$And take the derivative the same way as before: $27c^{2}x^{4}e^{-cx^{3}}$

b)

1. Apply Operator: $\dfrac{d(\cos(4ax^2))}{dx}$Take derivative with respect to cosine first, and then with respect to the function within the cosine: $-8ax\sin(4ax^{2})$
2. Apply Operator: $\dfrac{d^2(\cos(4ax^2))}{dx^2}$Which is the same as: $\dfrac{d(\dfrac{d(\cos(4ax^2))}{dx})}{dx}$Or: $\dfrac{d(-8ax\sin(4ax^{2}))}{dx}$ And take the derivative the same way as before, except now you need the Chain Rule. Take the derivative of the 'x' term before the sine and multiply by the sine function. Then take the derivative of the sine function and multiply by the 'x' function: $-8a\sin(4x^{2})-64a^{2}x^{2}\cos(4ax^{2})$

c)

1. Apply Operator: $\dfrac{d(e^{i^2kx^3})}{dx}$Take derivative of the whole function by multiplying the exponential by the derivative of its superscript. Note that the 'i' within the superscript refers to the imaginary number 'i':$-3kx^{2}e^{-kx^{3}}$
2. Apply Operator: $\dfrac{d^2(e^{i^2kx^3})}{dx^2}$Which is the same as: $\dfrac{d(-3kx^{2}e^{-kx^{3}})}{dx}$And take the derivative the same way as before, except now you need the Chain Rule. Take the derivative of the 'x' term before the exponential and multiply by the exponential function. Then take the derivative of the exponential function and multiply by the 'x' function: $-6kxe^{-kx^{3}}+9k^2x^4e^{-kx^3}$And finally, simplify:$3kxe^{-kx^{3}}(3kx^3-2)$

## 4.8: Particle in a One-Dimensional Box

### Q4.8.3

(a)

Solve by using the following general equation for the expectation value for energy for a particle in a 1D box:

$E_n = \dfrac{n^2 h^2}{8 m L^2}$

Solve for n = 1, n = 2, and n = 3 where $$h = 6.626\times 10^{-34} \;J \cdot s$$, $$m = 9.109\times 10^{-31} \;kg$$, and $$L = 5.0\times 10^{-10} \;m$$:

$E_1 = \dfrac{(1)^2 (6.626\times 10^{-34}\;J-s)^2}{8 (9.109\times 10^{-31}\;kg) (5.0\times 10^{-10}\;m)^2} = 2.41\times 10^{-19} \;J$

$E_3 = \dfrac{(3)^2 (6.626\times 10^{-34}\;J-s)^2}{8 (9.109\times 10^{-31}\;kg) (5.0\times 10^{-10}\;m)^2} = 2.17\times 10^{-18} \;J$

$E_5 = \dfrac{(5)^2 (6.626\times 10^{-34}\;J-s)^2}{8 (9.109\times 10^{-31}\;kg) (5.0\times 10^{-10}\;m)^2} = 6.02\times 10^{-18} \;J$

(b)

Find the energy emitted by finding the difference between E3 and E1:

$E_3 - E_1 = 2.17\times 10^{-18} \;J - 2.41\times 10^{-19} \;J = 1.93\times 10^{-18}\;J$

Find the wavelength of the energy emitted by the Planck-Einstein relation:

$\Delta E = h\nu$

$\Delta E = h(\dfrac{c}{\lambda})$

$\lambda = 1.03\times 10^{-7}\;m$

### Q4.8.4

Use Energy Equation:

$E_{n}=\dfrac{n^{2}h^{2}}{8mL^{2}}$

Substitute Average Energy from Question for En

$\dfrac{3}{2} \times kT=\dfrac{n^{2}h^{2}}{8mL^{2}}$

Solve this equation for n:

$n=\sqrt{\dfrac{12kTmL^{2}}{h^{2}}}(1)$

Convert 1 nm to _ m

$L=10^{-9} m$

Convert each temperature from oC to K

Then plug in the numbers into the Equation (1) to find the quantum number for the different temperatures:

$T=173K$

$n=\sqrt{\dfrac{12 \times (1.38 \times 10^{-23}J/K) \times 173K \times 6.65 \times 10^{-27}kg \times atom^{-1} \times 1^{-18}m^{2}}{(6.626 \times 10^{-34})^{2}Js}}$

$n=21$

$T=273K$

$n=\sqrt{\dfrac{12 \times (1.38 \times 10^{-23}J/K) \times 273K \times 6.65 \times 10^{-27}kg \times atom^{-1} \times 1^{-18}m^{2}}{(6.626 \times 10^{-34})^{2}Js}}$

$n=26$

$T=373K$

$n=\sqrt{\dfrac{12 \times (1.38 \times 10^{-23}J/K) \times 373K \times 6.65 \times 10^{-27}kg \times atom^{-1} \times 1^{-18}m^{2}}{(6.626 \times 10^{-34})^{2}Js}}$

$n=31$

## 4.10: The Schrödinger Wave Equation for the Hydrogen Atom

### Q4.10.1

For n = 4, l can have values of 0, 1, 2, and 3. Thus, s, p, d, and f subshells are found in the n = 4 shell of an atom. For l = 0 (the s subshell), ml can only be 0. Thus, there is only one 4s orbital. For l = 1 (p-type orbitals), m can have values of –1, 0, +1, so we find three 4p orbitals. For l = 2 (d-type orbitals), ml can have values of –2, –1, 0, +1, +2, so we have five 4d orbitals. When l = 3 (f-type orbitals), ml can have values of –3, –2, –1, 0, +1, +2, +3, and we can have seven 4f orbitals. Thus, we find a total of 16 orbitals in the n = 4 shell of an atom.

### Q4.10.2

(a) 3p (b) 5f (c) 2s

### Q4.10.3

(a) When n = 2, there are four orbitals (a single 2s orbital, and three orbitals labeled 2p). These four orbitals can contain eight electrons.

(b) When n = 5, there are five subshells of orbitals that we need to sum:

\begin{align} &\phantom{+}\textrm{1 orbitals labeled }5s\\ &\phantom{+}\textrm{3 orbitals labeled }5p\\ &\phantom{+}\textrm{5 orbitals labeled }5d\\ &\phantom{+}\textrm{7 orbitals labeled }5f\\ &\underline{+\textrm{9 orbitals labeled }5g}\\ &\,\textrm{25 orbitals total} \end{align}

Again, each orbital holds two electrons, so 50 electrons can fit in this shell.

(c) The number of orbitals in any shell n will equal n2. There can be up to two electrons in each orbital, so the maximum number of electrons will be 2 × n2

### Q4.10.4

The five degenerate 3d orbitals

## 4.11: Many-Electron Atoms & the Periodic Table

Exercises Solutions is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.