INT Model an atom as an electron in a rigid box of length 0.10...

Question

INT Model an atom as an electron in a rigid box of length 0.100 nm, roughly twice the Bohr radius. Calculate all the wavelengths that would be seen in the emission spectrum of this atom due to quantum jumps between these four energy levels. Give each wavelength a label λ_n→m to indicate the transition.

Accepted Answer

Hey everyone. So this problem is asking us to identify the wavelengths in the atoms emission spectrum due to quantum shifts between its three lowest energy states. Asked to consider the atom as an electron trapped in a one dimensional box with a length of 0.159 nanometers. We're also asked to label each wavelength as the wavelength or lambda of N to M to indicate the spec specific transitions. And so looking at the three lowest energy states are possible shifts are the wavelength from quantum number two to quantum number one, the wavelength from quantum number three to quantum number one or the wavelength from quantum number three to quantum number two. And so in that order in units of nanometers, our multiple choice answers are a 2810 and 20 B 1027 and 17 C 1710 and 27 or 2810 and 17. So the key to solving this problem is we're calling two different equations. So first, we have the equation for our wavelength and that is equal to H planks constant multiplied by C the speed of light divided by E are total energy. And then that energy of a particle is given by the equation N squared H squared or N is a quantum number all divided by eight multiplied by M or mass multiplied by L squared where L is the length of that box. And so as we transition between those at lower energy states, we are transitioning and changing between those quantum numbers. So we can write our change in energy from N to M as that change in quantum number where everything else stays the same because the particle isn't changing. So the mass isn't changing and the length of the box stays the same as well. So we have H squared divided by eight multiplied by M multiplied by L squared and all of that multiplied by N squared minus M squared to capture those two different energy levels. And so from there, we can find out our energies as the particles transition for each of those three shifts. And then finally solve, finally use that to solve for our wavelength. So what that looks like is we have our delta E is from and um is equal to H squared planks constant. We can recall 6.63 times 10 to the negative 34 dual seconds that quantity square divided by eight multiplied by the mass. We're told to assume it's an electron. So you can recall the mass of an electron is 9.11 times 10 to the negative 31 kg multiplied by that length of the box, which was given as 0.159 nanometers. So keeping everything in standard units, I'll rewrite that as 0.159 times 10 to the negative 9 m. And that quantity is square, all of that is multiplied by our N squared minus N squared. And so we can simplify and solve out for those known values. And we are left with delta E is equal to 2.39 T 10 to the negative 18 tools multiplied by N squared minus M squared. In turn, we can then use that to solve for our wavelength or lambda where we have lambda is equal to HC divided by E. So H again flanks constant 6.63 times 10 to the negative 34 dual seconds multiplied by the speed of light. We can recall is a constant three times 10 to the 8 m per second divided by that energy that we just solve for 2.39 times 10 to the negative 18 joules multiplied by N squared minus M squared. And so that's going to be the wavelength as we shift from quantum number N to quantum number N. When we plug those known values into our calculator, we get a wavelength in terms of N and M of 8.32 times 10 to the negative 8 m divided by N squared minus X squared. And so now it is a pretty simple problem to solve for those three energy shifts. So first, we have our wavelength from 2 to 1. And so we have 8.32 times 10 to the negative 8 m divided by two squared minus one square, which equals 2.8 times 10 to the negative 8 m which of course is equal to 28 nanometers. Likewise, or our wavelength from 3 to 1, we have 8.32 times 10 to the negative 8 m divided by three squared minus one squared. And that equals 1.0 times 10 to the negative 8 m or 10 nanometers. And then lastly from 32, our wavelength is equal to 8.32 times 10 to the negative 8 m divided by three squared minus two squared, which equals 1.7 times times the negative 8 m or 17 nanometers. And so our final answer is 28 nanometers, 10 nanometers and 17 nanometers. And when we look at our multiple choice answers, we see that, that aligns with answer choice D so D is the correct answer for this problem. That's all we have for this one. We'll see you in the next video.

Key Concepts

Quantum Mechanics

Particle in a Box Model

Emission Spectrum

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