(a) Find the excitation energy from the ground level to the third excited level for an electron confined to a box of width 0.360 0.360 nm. (b) The electron makes a transition from the n = 1 n = 1 to n = 4 n = 4 level by absorbing a photon. Calculate the wavelength of this photon.

Hey everyone. So this problem is dealing with quantum mechanics. Let's see what it's asking us. We have a containment chamber that is 0.470 nanometers wide. We're asked to determine the energy required to excite an electron from the second excited state to the fourth excited state. And part two is if a single photon, if a single photon has sufficient energy to excite the electron to make the transition in part one, what is the wavelength of that photon? Our multiple choice answers are a 6.82 electron volts and 182 nanometers b 27.3 electron volts and 45.5 nanometers C 8.42 times 10 to the negative 10 electron volts and 100 sorry 1470 m or D 3.37 times 10 to the negative nine electron volts and 336 m. So to solve the first part of this problem, we can recall that for the particle in a box model, our energy level equation is given by E sub N is equal to N squared which N where N is our level multiplied by H squared where H is planks constant, all divided by eight, multiplied by M, our mass multiplied by L squared where L is our length or in this case case, our width of this containment chamber. Now we are asked to solve for our change in energy between two states. So the fourth excited state is when N equals five. So that will be E of five. And then the second excited state is when N equals three. So E of five minus E sub three plugging in our N values, we get five squared multiplied by eight squared divided by eight ML squared minus three squared multiplied by that same term H squared divided by eight ML squared. And that allows us to simplify this equation where we have five squared minus three squared is 16. And so our delta E is simply 16 multiplied by H squared divided by eight multiplied by ML squared. And then that simplifies again to two. And so when we plug in our known values and our constants, we have two multiplied by H planks constant. We can recall is 6.63 times 10 to the negative 34 Joel seconds that quantity squared divided by our mass. So the mass of an electron we can recall is 9.11 times 10 to the negative 31 kg. And then our length is given as 0.47 nanometers. So keeping that in standard units, all right, 0.47 times 10 to the negative 9 m. And then that length is squared. So when we plug that in to our calcul calculators, we get 4.37 times 10 to the negative 18 jewels. But the problem is asking for it in units of electron rolls. So the last step is recalling our conversion factor where one ev is equal to 1.6 times 10 to the negative 19 joules. And that gives us 27.3 electron volts. And so looking at our multiple choice answers, the only correct answer for part one is B but we will solve her part two just to make sure that we're on the right track. So part two is asking for the wavelength for a photon of the same energy. And so we can recall that the energy of a photon is given by the equation E equals HC divided by lambda rearranging it for lambda or wavelength, we get the equation lambda equals HC divided by E where H is again, our planks constant. So 6.63 times 10 to the negative 34 J seconds multiplied by C our speed of light. We can recall is six times 10 to the 8 m per second, all divided by our energy. We'll be using the joules term um based off of our units here. So 4.37 times 10 to the negative 18 joules, we plug that in and we get 4.55 times 10 to the negative 8 m. And so just a little bit of unit manipulation here that equals 45.5 nanometers or 10, excuse me, 10 to the negative 9 m. And so looking back up at our multiple choice answers, we do confirm that B is the correct answer for this problem. So that's all we have for this one. We'll see you in the next video.

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Problem 18

Textbook Question

(a) Find the excitation energy from the ground level to the third excited level for an electron confined to a box of width $0.360$ nm.
(b) The electron makes a transition from the $n equals 1$ to $n equals 4$ level by absorbing a photon. Calculate the wavelength of this photon.

Verified step by step guidance

Step 1: Understand the energy levels of an electron in a one-dimensional box. The energy levels are given by the formula:

E = \frac{n^{2}}{h_{2}} 8 m L^{2}

, where

n

is the quantum number,

h_{2}

is Planck's constant squared,

m

is the mass of the electron, and

L

is the width of the box. For part (a), calculate the energy difference between the ground state (

n = 1

) and the third excited state (

n = 4

Step 2: Write the expression for the excitation energy:

ΔE = E n_{4} - E n_{1}

. Substitute the energy formula for each level:

ΔE = \frac{4^{2}}{h_{2}} 8 m L^{2} - \frac{1^{2}}{h_{2}} 8 m L^{2}

. Simplify the expression to find

ΔE

Step 3: For part (b), use the relationship between the energy of a photon and its wavelength:

E = \frac{hc}{λ}

, where

h

is Planck's constant,

c

is the speed of light, and

λ

is the wavelength of the photon. The energy of the photon corresponds to the energy difference between the

n = 4

and

n = 1

levels.

Step 4: Substitute the energy difference

ΔE

from part (a) into the photon energy formula to solve for the wavelength:

λ = \frac{hc}{ΔE}

. Ensure all constants are in consistent units (e.g., Planck's constant in Joule-seconds, speed of light in meters per second, and energy in Joules).

Step 5: Perform the necessary unit conversions for the box width (from nanometers to meters) and calculate the numerical values for the energy levels, energy difference, and wavelength. This will give you the wavelength of the photon absorbed during the transition.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Quantum Mechanics and Particle in a Box

In quantum mechanics, a particle confined in a one-dimensional box exhibits quantized energy levels. The energy levels are determined by the width of the box and the mass of the particle. For an electron in a box, the energy levels can be calculated using the formula E_n = (n^2 * h^2) / (8 * m * L^2), where n is the quantum number, h is Planck's constant, m is the mass of the electron, and L is the width of the box.

Energy Transitions and Photon Absorption

When an electron transitions between energy levels, it can absorb or emit a photon. The energy of the photon corresponds to the difference in energy between the two levels, given by ΔE = E_final - E_initial. This relationship is crucial for calculating the wavelength of the absorbed photon using the equation E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength.

Wavelength Calculation

The wavelength of a photon can be calculated from its energy using the equation λ = hc/E. This formula shows the inverse relationship between energy and wavelength: as the energy of the photon increases, its wavelength decreases. In the context of electron transitions, this calculation allows us to determine the specific wavelength of light absorbed when an electron moves from a lower to a higher energy level.

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(b) Since the energy in part (a) is all kinetic, to what speed does this correspond? How much time would it take at this speed for the ball to move from one side of the table to the other?

Textbook Question

A particle in a box is a billiard ball ( $m = 0.20$ kg) and the box has a width of $1.3$ m, the size of a billiard table. (Assume that the billiard ball slides without friction rather than rolls; that is, ignore the rotational kinetic energy.) What is the difference in energy between the $n equals 2$ and $n equals 1$ levels? Are quantum mechanical effects important for the game of billiards?

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Textbook Question

An electron in a one-dimensional box has ground state energy $2.00$ eV. What is the wavelength of the photon absorbed when the electron makes a transition to the second excited state?

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An electron is in a box of width $3.0 \times 10^{- 10}$ m. What are the de Broglie wavelength and the magnitude of the momentum of the electron if it is in (a) the $n equals 1$ level; (b) the $n equals 2$ level; (c) the $n equals 3$ level? In each case how does the wavelength compare to the width of the box?

Textbook Question

An electron is in a box of width 3.0*10^-10 m. What are the de Broglie wavelength and the magnitude of the momentum of the electron if it is in (a) the n = 1 level; (b) the n = 2 level; (c) the n = 3 level? In each case how does the wavelength compare to the width of the box?

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(a) An electron with initial kinetic energy $32$ eV encounters a square barrier with height $41$ eV and width $0.25$ nm. What is the probability that the electron will tunnel through the barrier?

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Textbook Question

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(a) Find the excitation energy from the ground level to the third excited level for an electron confined to a box of width 0.3600.360 nm.(b) The electron makes a transition from the n=1n = 1 to n=4n = 4 level by absorbing a photon. Calculate the wave­length of this photon.

Key Concepts

Quantum Mechanics and Particle in a Box

Energy Transitions and Photon Absorption

Wavelength Calculation

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(a) Find the excitation energy from the ground level to the third excited level for an electron confined to a box of width $0.360$ nm.
(b) The electron makes a transition from the $n equals 1$ to $n equals 4$ level by absorbing a photon. Calculate the wavelength of this photon.