Question 1

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)

Accepted Answer

The probability of tunneling can be calculated using the formula for the transmission coefficient T ≈ e^(-2 * k * a), where k = sqrt((2 * m * (V - E)) / ħ^2), m is the electron mass, V is the barrier height, E is the electron energy, a is the barrier width, and ħ is the reduced Planck's constant. Substituting the given values, the probability is approximately 1.4 × 10^-2.

Question 2

The energy of a particle in the second EXCITED state of a harmonic oscillator potential is What is the classical angular frequency of oscillation of this particle? (1 eV = 1.60 × 10^-19 J, = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The answer of The energy of a particle in the...

Question 3

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)

Accepted Answer

The ground state energy of an electron in an infinite well is given by:$$E_1 = \frac{h^2}{8mL^2}$$where:* $E_1$ is the ground state energy* $h$ is Planck's constant* $m$ is the mass of the electron* $L$ is the length of the boxSolving for $L$, we get:$$L = \sqrt{\frac{h^2}{8mE_1}}$$Substituting 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 4

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, = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The answer of An 80-eV electron impinges upon a potential...

Question 5

A lithium atom, mass 1.17 × 10-⁻²⁶ kg, vibrates with simple harmonic motion in a crystal lattice, where the effective force constant of the forces on the atom is k = 49.0 N/m. (c = 3.00 × 10⁸ m/s, h = 6.626 × 10⁻³⁴ J ∙ s, = 1.055 × 10⁻³⁴ J ∙ s, 1 eV = 1.60 × 10⁻¹⁹ J) (a) What is the ground state energy of this system, in eV? (b) What is the wavelength of the photon that could excite this system from the ground state to the first excited state?

Accepted Answer

The answer of A lithium atom, mass 1.17 × 10-⁻²⁶...

Question 6

An electron is confined in a harmonic oscillator potential well. What is the longest wavelength of light that the electron can absorb if the net force on the electron behaves as though it has a spring constant of 74 N/m? (^mel = 9.11 × 10^-31 kg, c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10^-19 J, = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The longest wavelength of light that the electron can absorb corresponds to the energy difference between two consecutive levels of the harmonic oscillator potential. The energy of a quantum harmonic oscillator is given by:

$E_n = \left(n + \frac{1}{2}ight)h
u$

where $n$ is the quantum number, $h$ is Planck's constant, and $
u$ is the frequency of the oscillator. The frequency of the oscillator is related to its spring constant $k$ by:

$
u = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where $m$ is the mass of the oscillator. In our case, the oscillator is an electron confined in a potential well, so $m$ is the electron mass. We are given that the net force on the electron behaves as though it has a spring constant of 74 N/m, so we can use this value to calculate the frequency of the oscillator:

$
u = \frac{1}{2\pi}\sqrt{\frac{74}{9.11	imes10^{-31}}} = 5.68	imes10^{15} \; 	ext{Hz}$

The energy difference between two consecutive levels of the oscillator is given by:

$\Delta E = E_{n+1} - E_n = h
u$

We want to find the longest wavelength of light that the electron can absorb, so we need to find the smallest energy difference. The smallest energy difference corresponds to the transition between the ground state ($n=0$) and the first excited state ($n=1$):

$\Delta E = E_1 - E_0 = \left(1 + \frac{1}{2}ight)h
u - \frac{1}{2}h
u = \frac{1}{2}h
u$

Substituting the value of $
u$ we found earlier, we get:

$\Delta E = \frac{1}{2}h
u = 4.24	imes10^{-20} \; 	ext{J}$

We can convert this energy to a wavelength using the formula:

$E = hc/\lambda$

where $\lambda$ is the wavelength of the light. Rearranging, we get:

$\lambda = hc/E$

Substituting the value of $\Delta E$ we found earlier, we get:

$\lambda = \frac{hc}{\Delta E} = \frac{hc}{4.24	imes10^{-20}} = 210 \; 	ext{nm}$

Therefore, the longest wavelength of light that the electron can absorb is 210 nm. The answer is A).

Question 7

A particle confined in a rigid one-dimensional box (an infinite well) of length 17.0 fm has an energy level E_n = 24.0 MeV 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)

Accepted Answer

The energy levels of a particle confined in a one-dimensional box (an infinite well) can be calculated using the equation: E_n = (n2ℏ2π2)/(2mL2) where n is the quantum number of the energy level, ℏ is the reduced Planck's constant, m is the mass of the particle, and L is the length of the box. We can use this equation to find the difference between the energy levels given: E_n+1 - E_n = [(n+1)2ℏ2π2/(2mL2)] - [n2ℏ2π2/(2mL2)] Simplifying this expression, we get: E_n+1 - E_n = ℏ2π2/(2mL2)[2n+1] We know the values of E_n+1, E_n, and L. We also know that ℏ = 1.055 × 10-34 J·s and that 1 MeV = 1.60 × 10-13 J. Using these values, we can solve for m: (37.5 MeV - 24.0 MeV) = (1.055 × 10-34 J·s)2π2/(2m(17.0 × 10-15 m)2)[2n+1] (13.5 × 1.60 × 10-13 J) = (1.055 × 10-34 J·s)2π2/(2m(17.0 × 10-15 m)2)[2n+1] m = 1.67 × 10-27 kg (mass of a proton) Now we can use the equation for the ground state energy level (n = 1): E_n = (ℏ2π2)/(2mL2) E_n = (1.055 × 10-34 J·s)2π2/(2(1.67 × 10-27 kg)(17.0 × 10-15 m)2) E_n = 1.50 MeV Therefore, the value of the ground state energy is 1.50 MeV, which is choice A.

Question 8

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, = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The answer of The energy of a proton is 1.0...

Question 9

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)

Accepted Answer

The answer of The wave function of an electron in...

Question 10

An electron is trapped in an infinite square well (a box) of width 6.88 nm. 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, ^mel = 9.11 × 10^-31 kg)

Accepted Answer

The energy of an electron in an infinite square well is 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 electron* $L$ is the width of the wellThe wavelength of a photon emitted when an electron drops from a higher energy state to a lower energy state is given by:$$\lambda = \frac{hc}{E_i - E_f}$$where:* $h$ is Planck's constant* $c$ is the speed of light* $E_i$ is the initial energy of the electron* $E_f$ is the final energy of the electronIn this problem, the electron drops from the $n = 5$ state to the $n = 1$ state. The energy of the electron in the $n = 5$ state is:$$E_5 = \frac{5^2h^2}{8mL^2}$$The energy of the electron in the $n = 1$ state is:$$E_1 = \frac{1^2h^2}{8mL^2}$$The wavelength of the photon emitted is:$$\lambda = \frac{hc}{E_5 - E_1} = \frac{hc}{4\frac{h^2}{8mL^2}} = \frac{2mL^2c}{h}$$Substituting the given values into the equation, we get:$$\lambda = \frac{2(9.11 	imes 10^{-31} 	ext{ kg})(6.88 	imes 10^{-9} 	ext{ m})^2(3.00 	imes 10^8 	ext{ m/s})}{6.626 	imes 10^{-34} 	ext{ J s}} = 6.49 	imes 10^{-6} 	ext{ m} = 6.49 \mu	ext{m}$$Therefore, the wavelength of the photon emitted is 6.49 μm.

Question 11

The lowest energy level of a certain quantum harmonic oscillator is 5.00 eV. What is the energy of the next higher level?

Accepted Answer

The answer of The lowest energy level of a certain...

Question 12

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)

Accepted Answer

The answer of You want to have an electron in...

Question 13

Calculate the ground state energy of a harmonic oscillator with a classical frequency of 3.68 × 10¹⁵ Hz. (1 eV = 1.60 × 10^-19 J, h = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The ground state energy of a harmonic oscillator is given by:E = (1/2)hfwhere h is Planck's constant and f is the classical frequency.Plugging in the given values, we get:E = (1/2)(1.055 × 10^-34 J s)(3.68 × 10^15 Hz) = 7.62 eVTherefore, the correct answer is A.

Question 14

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, h = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

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:$$\kappa = \sqrt{\frac{2m(E-V)}{\hbar^2}}$$where $m$ is the mass of the electron, $E$ is the energy of the electron, $V$ is the height of the barrier, and $\hbar$ is the reduced Planck constant.Plugging in the given values, we get:$$\kappa = \sqrt{\frac{2(9.11 	imes 10^{-31} 	ext{ kg})(3.10 	ext{ eV}-5.67 	ext{ eV})}{(1.055 	imes 10^{-34} 	ext{ J s})^2}} = 1.05 	imes 10^{10} 	ext{ m}^{-1}$$The width of the barrier is given as $0.40 	ext{ nm} = 4.00 	imes 10^{-10} 	ext{ m}$.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 15

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)

Accepted Answer

The minimum kinetic energy of a particle in an infinite potential well is given by the ground state energy, E = (h^2)/(8mL^2). Substituting the given values (h = 6.626 × 10^-34 J∙s, m = 10.0 g = 0.01 kg, L = 8.3 cm = 0.083 m) into this formula yields approximately 8.0 × 10^-64 J.

Question 16

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)

Accepted Answer

The energy levels of an electron in a one-dimensional box are given by:$$E_n = \frac{n^2h^2}{8mL^2}$$where:* $E_n$ is the energy of the electron in the nth energy level* $h$ is Planck's constant* $m$ is the mass of the electron* $L$ is the length of the boxThe difference in energy between two adjacent energy levels is given by:$$\Delta E = E_{n+1} - E_n = \frac{(n+1)^2h^2}{8mL^2} - \frac{n^2h^2}{8mL^2} = \frac{h^2}{8mL^2}$$We are given that the difference in energy between two adjacent energy levels is 1.068 × 10^-18 J and 1.352 × 10^-18 J. We can use this to solve for the length of the box:$$\Delta E = \frac{h^2}{8mL^2}$$$$L = \sqrt{\frac{h^2}{8m\Delta E}}$$$$L = \sqrt{\frac{(6.626 imes 10^{-34} ext{ J s})^2}{8(9.11 imes 10^{-31} ext{ kg})(1.068 imes 10^{-18} ext{ J})}}$$$$L = 1.9 ext{ nm}$$Therefore, the length of the box is 1.9 nm.

Question 17

An electron is confined in a harmonic oscillator potential well. A photon is emitted when the electron undergoes a 3→1 quantum jump. What is the wavelength of the emission if the net force on the electron behaves as though it has a spring constant of 9.6 N/m? (^mel = 9.11 × 10^-31kg, c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10^-19J, = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The answer of An electron is confined in a harmonic...

Question 18

The atoms in a nickel crystal vibrate as harmonic oscillators with an angular frequency of 2.3 × 10¹³ rad/s. The mass of a nickel atom is 9.75 × 10^-26 kg. What is the difference in energy between adjacent vibrational energy levels of nickel? (h = 6.626 × 10-³⁴ J ∙ s, = 1.055 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J)

Accepted Answer

The answer of The atoms in a nickel crystal vibrate...

Question 19

Find the wavelength of the photon emitted during the transition from the second EXCITED state to the ground state in a harmonic oscillator with a classical frequency of 3.72 × 10¹³ Hz. (c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10^-19 J, = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The answer of Find the wavelength of the photon emitted...

Question 20

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, ^mel = 9.11 × 10^-31 kg)

Accepted Answer

The energy of a photon is given by E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength. The energy difference between the n = 2 and n = 1 states is given by ΔE = E2 - E1 = (hc/λ2) - (hc/λ1). The width of the molecule is given by L = √(2mΔE/h^2), where m is the mass of the electron. Substituting the given values, we get L = 1.33 nm.

Quiz 39: Quantum Mechanics

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)

The energy of a particle in the second EXCITED state of a harmonic oscillator potential is What is the classical angular frequency of oscillation of this particle? (1 eV = 1.60 × 10^-19 J, = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

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)

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, = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

A particle confined in a rigid one-dimensional box (an infinite well) of length 17.0 fm has an energy level E_n = 24.0 MeV 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)

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, = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

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)

An electron is trapped in an infinite square well (a box) of width 6.88 nm. 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, ^mel = 9.11 × 10^-31 kg)

The lowest energy level of a certain quantum harmonic oscillator is 5.00 eV. What is the energy of the next higher level?

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)

Calculate the ground state energy of a harmonic oscillator with a classical frequency of 3.68 × 10¹⁵ Hz. (1 eV = 1.60 × 10^-19 J, h = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

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, h = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

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)

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)

Find the wavelength of the photon emitted during the transition from the second EXCITED state to the ground state in a harmonic oscillator with a classical frequency of 3.72 × 10¹³ Hz. (c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10^-19 J, = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

Quiz 39: Quantum Mechanics

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)

The energy of a particle in the second EXCITED state of a harmonic oscillator potential is What is the classical angular frequency of oscillation of this particle? (1 eV = 1.60 × 10-19 J, = 1.055 × 10-34 J ∙ s, h = 6.626 × 10-34 J ∙ s)

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)

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, = 1.055 × 10-34 J ∙ s, h = 6.626 × 10-34 J ∙ s)

A particle confined in a rigid one-dimensional box (an infinite well) of length 17.0 fm has an energy level En = 24.0 MeV 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)

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, = 1.055 × 10-34 J ∙ s, h = 6.626 × 10-34 J ∙ s)

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)

An electron is trapped in an infinite square well (a box) of width 6.88 nm. 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, mel = 9.11 × 10-31 kg)

The lowest energy level of a certain quantum harmonic oscillator is 5.00 eV. What is the energy of the next higher level?

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)

Calculate the ground state energy of a harmonic oscillator with a classical frequency of 3.68 × 1015 Hz. (1 eV = 1.60 × 10-19 J, h = 1.055 × 10-34 J ∙ s, h = 6.626 × 10-34 J ∙ s)

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, h = 1.055 × 10-34 J ∙ s, h = 6.626 × 10-34 J ∙ s)

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)

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)

Find the wavelength of the photon emitted during the transition from the second EXCITED state to the ground state in a harmonic oscillator with a classical frequency of 3.72 × 1013 Hz. (c = 3.00 × 108 m/s, 1 eV = 1.60 × 10-19 J, = 1.055 × 10-34 J ∙ s, h = 6.626 × 10-34 J ∙ s)

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)

The energy of a particle in the second EXCITED state of a harmonic oscillator potential is What is the classical angular frequency of oscillation of this particle? (1 eV = 1.60 × 10^-19 J, = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

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)

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, = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

A particle confined in a rigid one-dimensional box (an infinite well) of length 17.0 fm has an energy level E_n = 24.0 MeV 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)

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, = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

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)

An electron is trapped in an infinite square well (a box) of width 6.88 nm. 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, ^mel = 9.11 × 10^-31 kg)

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)

Calculate the ground state energy of a harmonic oscillator with a classical frequency of 3.68 × 10¹⁵ Hz. (1 eV = 1.60 × 10^-19 J, h = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

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, h = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

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)

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)

Find the wavelength of the photon emitted during the transition from the second EXCITED state to the ground state in a harmonic oscillator with a classical frequency of 3.72 × 10¹³ Hz. (c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10^-19 J, = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)