Question 1

Tunneling: A 3.10-eV electron is incident on a 0.40-nm barrier that is 5.67 eV high. What is the probability that this electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_el = 9.11 × 10^-31 kg, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s) A) 1.4 × 10^-3 B) 9.0 × 10^-4 C) 1.0 × 10^-3 D) 1.5 × 10^-3 E) 1.2 × 10^-3

Accepted Answer

The probability of tunneling is given by the formula:$$T = e^{-2\kappa L}$$where $\kappa$ is the wave vector of the electron in the barrier and $L$ is the width of the barrier.The wave vector is given by:$$\kappa = \sqrt{\frac{2m(V_0 - E)}{\hbar^2}}$$where $m$ is the mass of the electron, $V_0$ is the height of the barrier, and $E$ is the energy of the electron.Plugging in the values given in the problem, we get:$$\kappa = \sqrt{\frac{2(9.11 	imes 10^{-31} 	ext{ kg})(5.67 	ext{ eV} - 3.10 	ext{ eV})}{(1.055 	imes 10^{-34} 	ext{ J s})^2}} = 1.05 	imes 10^{10} 	ext{ m}^{-1}$$The width of the barrier is given as $L = 0.40 	ext{ nm} = 4.00 	imes 10^{-10} 	ext{ m}$.Plugging these values into the formula for the probability of tunneling, we get:$$T = e^{-2(1.05 	imes 10^{10} 	ext{ m}^{-1})(4.00 	imes 10^{-10} 	ext{ m})} = 1.4 	imes 10^{-3}$$Therefore, the probability that the electron will tunnel through the barrier is $1.4 	imes 10^{-3}$.

Question 2

Particle in a box: You want to have an electron in an energy level where its speed is no more than 66 m/s. What is the length of the smallest box (an infinite well) in which you can do this? (h = 6.626 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg) A) 5.5 µm B) 11 µm C) 2.8 µm D) 1.4 µm

Accepted Answer

The de Broglie wavelength of the electron must be less than or equal to twice the length of the box. The de Broglie wavelength is given by h/p, where h is Planck's constant and p is the momentum of the electron. The momentum of the electron is given by mv, where m is the mass of the electron and v is its speed. Substituting these expressions into the de Broglie wavelength equation, we get:λ = h/mvSetting this equal to twice the length of the box, we get:2L = h/mvSolving for L, we get:L = h/2mvSubstituting the given values, we get:L = (6.626 × 10^-34 J ∙ s)/(2 × 9.11 × 10^-31 kg × 66 m/s) = 5.5 µmTherefore, the length of the smallest box in which you can do this is 5.5 µm.

Question 3

Particle in a box: A particle confined in a rigid one-dimensional box (an infinite well) of length 17.0 fm has an energy level and an adjacent energy level E_n₊₁ = 37.5 MeV. What is the value of the ground state energy? (1 eV = 1.60 × 10^-19 J) A) 1.50 MeV B) 13.5 MeV C) 0.500 MeV D) 4.50 MeV

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Question 4

Particle in a box: You want to confine an electron in a box (an infinite well) so that its ground state energy is 5.0 × 10^-18 J. What should be the length of the box? (h = 6.626 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg) A) 0.11 nm B) 0.22 nm C) 0.15 nm D) 0.18 nm

Accepted Answer

The ground state energy of a particle in a box is given by:$$E_1 = \frac{h^2}{8mL^2}$$where h is Planck's constant, m is the mass of the particle, and L is the length of the box.We can rearrange this equation to solve for L:$$L = \sqrt{\frac{h^2}{8mE_1}}$$Plugging in the given values, we get:$$L = \sqrt{\frac{(6.626 	imes 10^{-34} 	ext{ J s})^2}{8(9.11 	imes 10^{-31} 	ext{ kg})(5.0 	imes 10^{-18} 	ext{ J})}} = 0.11 	ext{ nm}$$Therefore, the length of the box should be 0.11 nm.

Question 5

Tunneling: An 80-eV electron impinges upon a potential barrier 100 eV high and 0.20 nm thick. What is the probability the electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_proton = 1.67 × 10^-27 kg, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s) A) 0.027% B) 2.7% C) 0.27% D) 2.8 × 10^-4 % E) 2.0 × 10^-9 %

Accepted Answer

The tunneling probability can be calculated using the formula: T ≈ e^(-2 * κ * a), where κ = sqrt((2 * m * (V - E)) / ħ²), m is the electron mass (9.11 × 10^-31 kg), V is the barrier height (100 eV), E is the electron energy (80 eV), and a is the barrier width (0.20 nm). After calculating κ and substituting the values, the tunneling probability is approximately 0.00027 or 0.027%.

Question 6

Tunneling: An electron with kinetic energy 2.80 eV encounters a potential barrier of height 4.70 eV. If the barrier width is 0.40 nm, what is the probability that the electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_el = 9.11 × 10^-31 kg, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s) A) 3.5 × 10^-3 B) 7.3 × 10^-3 C) 1.4 × 10^-2 D) 2.9 × 10^-3 E) 3.5 × 10^-2

Accepted Answer

The probability of tunneling through a barrier is given by the formula $ T \approx e^{-2 \kappa a} $, where $ \kappa = \sqrt{\frac{2m(V_0 - E)}{\hbar^2}} $. Substituting the given values, $ \kappa \approx 1.145 \times 10^{10} \, \text{m}^{-1} $ and $ a = 0.40 \, \text{nm} = 0.40 \times 10^{-9} \, \text{m} $, we find $ T \approx e^{-2 \times 1.145 \times 10^{10} \times 0.40 \times 10^{-9}} \approx 3.5 \times 10^{-3} $.

Question 7

Particle in a box: An electron is trapped in an infinite square well (a box) of width Find the wavelength of photons emitted when the electron drops from the n = 5 state to the n = 1 state in this system. (c = 3.00 × 10⁸ m/s, h = 6.626 × 10^-34J ∙ s, m_el = 9.11 × 10^-31 kg) A) 6.49 μm B) 5.45 μm C) 5.91 μm D) 7.07 μm

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Question 8

Particle in a box: One fairly crude method of determining the size of a molecule is to treat the molecule as an infinite square well (a box) with an electron trapped inside, and to measure the wavelengths of emitted photons. If the photon emitted during the n = 2 to n = 1 transition has wavelength 1940 nm, what is the width of the molecule? (c = 3.00 × 10⁸ m/s, h = 6.626 × 10^-34J ∙ s, m_el = 9.11 × 10^-31 kg) A) 1.33 nm B) 1.12 nm C) 1.21 nm D) 1.45 nm

Accepted Answer

The energy difference between the n = 2 and n = 1 states in an infinite square well is given by $\Delta E = \frac{h^2}{8mL^2}(n_2^2 - n_1^2)$, where $h$ is Planck's constant, $m$ is the mass of the electron, and $L$ is the width of the well. The wavelength of the emitted photon can be related to the energy difference by $\Delta E = \frac{hc}{\lambda}$, where $c$ is the speed of light and $\lambda$ is the wavelength of the photon. Substituting the given values and solving for $L$ yields approximately 1.33 nm.

Question 9

Particle in a box: A 10.0-g bouncy ball is confined in a 8.3-cm-long box (an infinite well). If we model the ball as a point particle, what is the minimum kinetic energy of the ball? (h = 6.626 × 10^-34 J ∙ s) A) 8.0 × J B) 3.2 × J C) 1.3 × J D) zero

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Question 10

Particle in a box: An electron is confined in a one-dimensional box (an infinite well). Two adjacent allowed energies of the electron are 1.068 × 10^-18 J and 1.352 × 10^-18 J. What is the length of the box? (h = 6.626 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg) A) 1.9 nm B) 0.93 nm C) 1.1 nm D) 2.3 nm

Accepted Answer

The energy levels of a particle in a one-dimensional box are given by:$$E_n = \frac{n^2h^2}{8mL^2}$$where n is the principal quantum number, h is Planck's constant, m is the mass of the particle, and L is the length of the box.The difference in energy between two adjacent levels is:$$\Delta E = E_{n+1} - E_n = \frac{(n+1)^2h^2}{8mL^2} - \frac{n^2h^2}{8mL^2} = \frac{2nh^2}{8mL^2}$$Substituting the given values, we get:$$\Delta E = 1.352 	imes 10^{-18} 	ext{ J} - 1.068 	imes 10^{-18} 	ext{ J} = 2.84 	imes 10^{-19} 	ext{ J}$$$$2.84 	imes 10^{-19} 	ext{ J} = \frac{2nh^2}{8mL^2}$$$$L^2 = \frac{2nh^2}{8m\Delta E}$$$$L = \sqrt{\frac{2nh^2}{8m\Delta E}}$$$$L = \sqrt{\frac{2(1)(6.626 	imes 10^{-34} 	ext{ J s})^2}{8(9.11 	imes 10^{-31} 	ext{ kg})(2.84 	imes 10^{-19} 	ext{ J})}}$$$$L = 1.9 	imes 10^{-9} 	ext{ m} = 1.9 	ext{ nm}$$Therefore, the length of the box is 1.9 nm.

Question 11

Particle in a box: The wave function of an electron in a rigid box (infinite well) is shown in the figure. If the electron energy 98.0 eV, what is the energy of the electron's ground state? ( m_el = 9.11 × 10^-31 kg) A) 6.13 eV B) 3.92 eV C) 10.9 eV D) 24.5 eV

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Question 12

Particle in a box: An electron is bound in an infinite well (a box) of width 0.10 nm. If the electron is initially in the n = 8 state and falls to the n = 7 state, find the wavelength of the emitted photon. (c = 3.00 × 10⁸ m/s, h = 6.626 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg)

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Question 13

Particle in a box: A one-dimensional finite potential well has potential energy U₀= 0 in the region 0 < x < 0.2 nm, and U₀= 15 eV everywhere else. A particle with which of the energies listed below would be localized (trapped) within the potential well? (Select all correct answers.) A) 5 eV B) 20 eV C) 16 eV D) 7 eV E) None of the above.

Accepted Answer

For a particle to be localized within a potential well, its energy must be less than the potential energy outside the well (otherwise it would escape) but greater than the potential energy inside the well (otherwise it would tunnel through the barrier). 
In this case, the potential energy inside the well is 0 eV and the potential energy outside the well is 15 eV. Therefore, the particle's energy must be between 0 and 15 eV.
Option A (5 eV) and Option D (7 eV) both fall within this range and hence, a particle with either of these energies would be localized within the potential well. 
Options B (20 eV) and C (16 eV) have energies greater than the potential energy outside the well and hence, the particle would not be trapped within the potential well. 
Therefore, the correct answer is choice A and D.

Question 14

Tunneling: The energy of a proton is 1.0 MeV below the top of a 6.8-fm-wide energy barrier. What is the probability that the proton will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_proton = 1.67 × 10^-27 kg, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s) A) 0.051 B) 0.048 C) 0.056 D) 0.053

Accepted Answer

The probability of tunneling is given by the formula:$$P = e^{-2\kappa L}$$where $\kappa$ is the wave number of the particle in the barrier and $L$ is the width of the barrier.The wave number is given by:$$\kappa = \sqrt{\frac{2m(V_0 - E)}{\hbar^2}}$$where $m$ is the mass of the particle, $V_0$ is the height of the barrier, $E$ is the energy of the particle, and $\hbar$ is the reduced Planck constant.Plugging in the given values, we get:$$\kappa = \sqrt{\frac{2(1.67 	imes 10^{-27} 	ext{ kg})(1.0 	ext{ MeV})(1.60 	imes 10^{-19} 	ext{ J/eV})}{(1.055 	imes 10^{-34} 	ext{ J s})^2}} = 1.05 	imes 10^{11} 	ext{ m}^{-1}$$The width of the barrier is given as $L = 6.8 	ext{ fm} = 6.8 	imes 10^{-15} 	ext{ m}$.Plugging these values into the formula for the probability of tunneling, we get:$$P = e^{-2(1.05 	imes 10^{11} 	ext{ m}^{-1})(6.8 	imes 10^{-15} 	ext{ m})} = 0.051$$Therefore, the probability that the proton will tunnel through the barrier is 0.051.

Question 15

Tunneling: An 80-eV electron impinges upon a potential barrier 100 eV high and 0.20 nm thick. What is the probability the electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_proton = 1.67 × 10^-27 kg, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s) A) 0.011% B) 1.1% C) 0.11% D) 1.1 × 10^-4 % E) 7.7 × 10^-10 %

Accepted Answer

The probability of tunneling can be calculated using the formula for the transmission coefficient T ≈ e^(-2 * κ * a), where κ = sqrt(2m(V-E))/ħ. Substituting the given values, the probability is approximately 0.011%.

Question 16

Tunneling: An electron with kinetic energy 2.80 eV encounters a potential barrier of height 4.70 eV. If the barrier width is 0.40 nm, what is the probability that the electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_el = 9.11 × 10^-31 kg, h = 6.626 × 10^-34 J ∙ s) A) 1.4 × 10^-2 B) 2.8 × 10^-2 C) 5.5 × 10^-2 D) 1.1 × 10^-2 E) 1.4 × 10^-1

Accepted Answer

The probability of tunneling can be calculated using the formula: T ≈ e^(-2 * k * a), where k = sqrt(2 * m * (V - E)) / ħ. Substituting the given values, we find that the probability is approximately 1.4 × 10^-2.

Question 17

Particle in a box: The lowest energy level of a particle confined to a one-dimensional region of space (a box, or infinite well) with fixed length L is E₀. If an identical particle is confined to a similar region with fixed length L/ 6, what is the energy of the lowest energy level that the particles have in common? Express your answer in terms of E₀.

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Question 18

Tunneling: The energy of a proton is 1.0 MeV below the top of a 1.2-MeV-high energy barrier that is 6.8 fm wide. What is the probability that the proton will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_proton = 1.67 × 10^-27 kg, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s) A) 11% B) 9.1% C) 14% D) 7.5%

Accepted Answer

The probability of tunneling is given by the formula:$$P = e^{-2\kappa L}$$where $\kappa$ is the wave number of the particle in the barrier and $L$ is the width of the barrier. The wave number is given by:$$\kappa = \sqrt{\frac{2m(E-V)}{\hbar^2}}$$where $m$ is the mass of the particle, $E$ is the energy of the particle, $V$ is the height of the barrier, and $\hbar$ is the reduced Planck constant.Plugging in the given values, we get:$$\kappa = \sqrt{\frac{2(1.67 	imes 10^{-27} 	ext{ kg})(1.0 	imes 10^6 	ext{ eV})(1.60 	imes 10^{-19} 	ext{ J/eV})}{(1.055 	imes 10^{-34} 	ext{ J s})^2}} = 1.05 	imes 10^{10} 	ext{ m}^{-1}$$$$P = e^{-2(1.05 	imes 10^{10} 	ext{ m}^{-1})(6.8 	imes 10^{-15} 	ext{ m})} = 0.11$$Therefore, the probability that the proton will tunnel through the barrier is 11%.

Question 19

Tunneling: A 3.10-eV electron is incident on a 0.40-nm barrier that is 5.67 eV high. What is the probability that this electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_el = 9.11 × 10^-31 kg, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s) A) 0.56% B) 0.35% C) 0.40% D) 0.25% E) 0.48%

Accepted Answer

The probability of tunneling is given by the following equation:$$T = e^{-2\kappa L}$$where $\kappa$ is the wave vector of the electron in the barrier and $L$ is the width of the barrier. The wave vector is given by:$$\kappa = \sqrt{\frac{2m(V_0 - E)}{\hbar^2}}$$where $m$ is the mass of the electron, $V_0$ is the height of the barrier, and $E$ is the energy of the electron.Plugging in the given values, we get:$$\kappa = \sqrt{\frac{2(9.11 	imes 10^{-31} 	ext{ kg})(5.67 	ext{ eV} - 3.10 	ext{ eV})}{(1.055 	imes 10^{-34} 	ext{ J s})^2}} = 1.05 	imes 10^{10} 	ext{ m}^{-1}$$The width of the barrier is given as $L = 0.40 	ext{ nm} = 4.00 	imes 10^{-10} 	ext{ m}$.Therefore, the probability of tunneling is:$$T = e^{-2(1.05 	imes 10^{10} 	ext{ m}^{-1})(4.00 	imes 10^{-10} 	ext{ m})} = 0.0056 = 0.56%$$

Question 20

Particle in a box: An electron in an infinite potential well (a box) makes a transition from the n = 3 level to the ground state and in so doing emits a photon of wavelength 20.9 nm. (c = 3.00 × 10⁸ m/s, h = 6.626 × 10^-34 J ∙ s, m_e_l = 9.11 × 10^-31kg) (a) What is the width of this well? (b) What wavelength photon would be required to excite the electron from its original level to the next higher one?

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Quiz 31: Alternating Current

Tunneling: A 3.10-eV electron is incident on a 0.40-nm barrier that is 5.67 eV high. What is the probability that this electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_el = 9.11 × 10^-31 kg, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

Particle in a box: You want to have an electron in an energy level where its speed is no more than 66 m/s. What is the length of the smallest box (an infinite well) in which you can do this? (h = 6.626 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg)

Particle in a box: A particle confined in a rigid one-dimensional box (an infinite well) of length 17.0 fm has an energy level and an adjacent energy level E_n₊₁ = 37.5 MeV. What is the value of the ground state energy? (1 eV = 1.60 × 10^-19 J)

Particle in a box: You want to confine an electron in a box (an infinite well) so that its ground state energy is 5.0 × 10^-18 J. What should be the length of the box? (h = 6.626 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg)

Tunneling: An 80-eV electron impinges upon a potential barrier 100 eV high and 0.20 nm thick. What is the probability the electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_proton = 1.67 × 10^-27 kg, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

Particle in a box: An electron is trapped in an infinite square well (a box) of width Find the wavelength of photons emitted when the electron drops from the n = 5 state to the n = 1 state in this system. (c = 3.00 × 10⁸ m/s, h = 6.626 × 10^-34J ∙ s, m_el = 9.11 × 10^-31 kg)

Particle in a box: A 10.0-g bouncy ball is confined in a 8.3-cm-long box (an infinite well) . If we model the ball as a point particle, what is the minimum kinetic energy of the ball? (h = 6.626 × 10^-34 J ∙ s)

Particle in a box: An electron is confined in a one-dimensional box (an infinite well) . Two adjacent allowed energies of the electron are 1.068 × 10^-18 J and 1.352 × 10^-18 J. What is the length of the box? (h = 6.626 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg)

Particle in a box: The wave function of an electron in a rigid box (infinite well) is shown in the figure. If the electron energy 98.0 eV, what is the energy of the electron's ground state? ( m_el = 9.11 × 10^-31 kg)

Particle in a box: An electron is bound in an infinite well (a box) of width 0.10 nm. If the electron is initially in the n = 8 state and falls to the n = 7 state, find the wavelength of the emitted photon. (c = 3.00 × 10⁸ m/s, h = 6.626 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg)

Particle in a box: A one-dimensional finite potential well has potential energy U₀= 0 in the region 0 < x < 0.2 nm, and U₀= 15 eV everywhere else. A particle with which of the energies listed below would be localized (trapped) within the potential well? (Select all correct answers.)

Tunneling: The energy of a proton is 1.0 MeV below the top of a 6.8-fm-wide energy barrier. What is the probability that the proton will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_proton = 1.67 × 10^-27 kg, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

Tunneling: An 80-eV electron impinges upon a potential barrier 100 eV high and 0.20 nm thick. What is the probability the electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_proton = 1.67 × 10^-27 kg, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

Tunneling: An electron with kinetic energy 2.80 eV encounters a potential barrier of height 4.70 eV. If the barrier width is 0.40 nm, what is the probability that the electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_el = 9.11 × 10^-31 kg, h = 6.626 × 10^-34 J ∙ s)

Tunneling: The energy of a proton is 1.0 MeV below the top of a 1.2-MeV-high energy barrier that is 6.8 fm wide. What is the probability that the proton will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_proton = 1.67 × 10^-27 kg, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

Tunneling: A 3.10-eV electron is incident on a 0.40-nm barrier that is 5.67 eV high. What is the probability that this electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_el = 9.11 × 10^-31 kg, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

Quiz 31: Alternating Current

Tunneling: A 3.10-eV electron is incident on a 0.40-nm barrier that is 5.67 eV high. What is the probability that this electron will tunnel through the barrier? (1 eV = 1.60 × 10-19 J, mel = 9.11 × 10-31 kg, ħ = 1.055 × 10-34 J ∙ s, h = 6.626 × 10-34 J ∙ s)

Particle in a box: You want to have an electron in an energy level where its speed is no more than 66 m/s. What is the length of the smallest box (an infinite well) in which you can do this? (h = 6.626 × 10-34 J ∙ s, mel = 9.11 × 10-31 kg)

Particle in a box: A particle confined in a rigid one-dimensional box (an infinite well) of length 17.0 fm has an energy level and an adjacent energy level En+1 = 37.5 MeV. What is the value of the ground state energy? (1 eV = 1.60 × 10-19 J)

Particle in a box: You want to confine an electron in a box (an infinite well) so that its ground state energy is 5.0 × 10-18 J. What should be the length of the box? (h = 6.626 × 10-34 J ∙ s, mel = 9.11 × 10-31 kg)

Tunneling: An 80-eV electron impinges upon a potential barrier 100 eV high and 0.20 nm thick. What is the probability the electron will tunnel through the barrier? (1 eV = 1.60 × 10-19 J, mproton = 1.67 × 10-27 kg, ħ = 1.055 × 10-34 J ∙ s, h = 6.626 × 10-34 J ∙ s)

Particle in a box: An electron is trapped in an infinite square well (a box) of width Find the wavelength of photons emitted when the electron drops from the n = 5 state to the n = 1 state in this system. (c = 3.00 × 108 m/s, h = 6.626 × 10-34 J ∙ s, mel = 9.11 × 10-31 kg)

Particle in a box: A 10.0-g bouncy ball is confined in a 8.3-cm-long box (an infinite well) . If we model the ball as a point particle, what is the minimum kinetic energy of the ball? (h = 6.626 × 10-34 J ∙ s)

Particle in a box: An electron is confined in a one-dimensional box (an infinite well) . Two adjacent allowed energies of the electron are 1.068 × 10-18 J and 1.352 × 10-18 J. What is the length of the box? (h = 6.626 × 10-34 J ∙ s, mel = 9.11 × 10-31 kg)

Particle in a box: The wave function of an electron in a rigid box (infinite well) is shown in the figure. If the electron energy 98.0 eV, what is the energy of the electron's ground state? ( mel = 9.11 × 10-31 kg)

Particle in a box: An electron is bound in an infinite well (a box) of width 0.10 nm. If the electron is initially in the n = 8 state and falls to the n = 7 state, find the wavelength of the emitted photon. (c = 3.00 × 108 m/s, h = 6.626 × 10-34 J ∙ s, mel = 9.11 × 10-31 kg)

Particle in a box: A one-dimensional finite potential well has potential energy U0 = 0 in the region 0 < x < 0.2 nm, and U0 = 15 eV everywhere else. A particle with which of the energies listed below would be localized (trapped) within the potential well? (Select all correct answers.)

Tunneling: The energy of a proton is 1.0 MeV below the top of a 6.8-fm-wide energy barrier. What is the probability that the proton will tunnel through the barrier? (1 eV = 1.60 × 10-19 J, mproton = 1.67 × 10-27 kg, ħ = 1.055 × 10-34 J ∙ s, h = 6.626 × 10-34 J ∙ s)

Tunneling: An 80-eV electron impinges upon a potential barrier 100 eV high and 0.20 nm thick. What is the probability the electron will tunnel through the barrier? (1 eV = 1.60 × 10-19 J, mproton = 1.67 × 10-27 kg, ħ = 1.055 × 10-34 J ∙ s, h = 6.626 × 10-34 J ∙ s)

Tunneling: An electron with kinetic energy 2.80 eV encounters a potential barrier of height 4.70 eV. If the barrier width is 0.40 nm, what is the probability that the electron will tunnel through the barrier? (1 eV = 1.60 × 10-19 J, mel = 9.11 × 10-31 kg, h = 6.626 × 10-34 J ∙ s)

Tunneling: The energy of a proton is 1.0 MeV below the top of a 1.2-MeV-high energy barrier that is 6.8 fm wide. What is the probability that the proton will tunnel through the barrier? (1 eV = 1.60 × 10-19 J, mproton = 1.67 × 10-27 kg, ħ = 1.055 × 10-34 J ∙ s, h = 6.626 × 10-34 J ∙ s)

Tunneling: A 3.10-eV electron is incident on a 0.40-nm barrier that is 5.67 eV high. What is the probability that this electron will tunnel through the barrier? (1 eV = 1.60 × 10-19 J, mel = 9.11 × 10-31 kg, ħ = 1.055 × 10-34 J ∙ s, h = 6.626 × 10-34 J ∙ s)

Tunneling: A 3.10-eV electron is incident on a 0.40-nm barrier that is 5.67 eV high. What is the probability that this electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_el = 9.11 × 10^-31 kg, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

Particle in a box: You want to have an electron in an energy level where its speed is no more than 66 m/s. What is the length of the smallest box (an infinite well) in which you can do this? (h = 6.626 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg)

Particle in a box: A particle confined in a rigid one-dimensional box (an infinite well) of length 17.0 fm has an energy level and an adjacent energy level E_n₊₁ = 37.5 MeV. What is the value of the ground state energy? (1 eV = 1.60 × 10^-19 J)

Particle in a box: You want to confine an electron in a box (an infinite well) so that its ground state energy is 5.0 × 10^-18 J. What should be the length of the box? (h = 6.626 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg)

Tunneling: An 80-eV electron impinges upon a potential barrier 100 eV high and 0.20 nm thick. What is the probability the electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_proton = 1.67 × 10^-27 kg, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

Particle in a box: An electron is trapped in an infinite square well (a box) of width Find the wavelength of photons emitted when the electron drops from the n = 5 state to the n = 1 state in this system. (c = 3.00 × 10⁸ m/s, h = 6.626 × 10^-34J ∙ s, m_el = 9.11 × 10^-31 kg)

Particle in a box: A 10.0-g bouncy ball is confined in a 8.3-cm-long box (an infinite well) . If we model the ball as a point particle, what is the minimum kinetic energy of the ball? (h = 6.626 × 10^-34 J ∙ s)

Particle in a box: An electron is confined in a one-dimensional box (an infinite well) . Two adjacent allowed energies of the electron are 1.068 × 10^-18 J and 1.352 × 10^-18 J. What is the length of the box? (h = 6.626 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg)

Particle in a box: The wave function of an electron in a rigid box (infinite well) is shown in the figure. If the electron energy 98.0 eV, what is the energy of the electron's ground state? ( m_el = 9.11 × 10^-31 kg)

Particle in a box: An electron is bound in an infinite well (a box) of width 0.10 nm. If the electron is initially in the n = 8 state and falls to the n = 7 state, find the wavelength of the emitted photon. (c = 3.00 × 10⁸ m/s, h = 6.626 × 10^-34 J ∙ s, m_el = 9.11 × 10^-31 kg)

Particle in a box: A one-dimensional finite potential well has potential energy U₀= 0 in the region 0 < x < 0.2 nm, and U₀= 15 eV everywhere else. A particle with which of the energies listed below would be localized (trapped) within the potential well? (Select all correct answers.)

Tunneling: The energy of a proton is 1.0 MeV below the top of a 6.8-fm-wide energy barrier. What is the probability that the proton will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_proton = 1.67 × 10^-27 kg, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

Tunneling: An 80-eV electron impinges upon a potential barrier 100 eV high and 0.20 nm thick. What is the probability the electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_proton = 1.67 × 10^-27 kg, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

Tunneling: An electron with kinetic energy 2.80 eV encounters a potential barrier of height 4.70 eV. If the barrier width is 0.40 nm, what is the probability that the electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_el = 9.11 × 10^-31 kg, h = 6.626 × 10^-34 J ∙ s)

Tunneling: A 3.10-eV electron is incident on a 0.40-nm barrier that is 5.67 eV high. What is the probability that this electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_el = 9.11 × 10^-31 kg, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)