The allowed energies of a simple atom are 0.00 eV, 4.00 eV, and 6...

Question

The allowed energies of a simple atom are 0.00 eV, 4.00 eV, and 6.00 eV.c. What wavelengths appear in the atom’s absorption spectrum?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the fine practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Compute the wavelengths present in the absorption spectrum of a hydrogen like atom with distinct energy levels at zero point 00 electron volts, 3.50 electron volts and 5.50 electron volts. So that's our end goals. We're trying to figure out what the wavelengths are present in the absorption spectrum of a hydrogen like atom that has three distinct energy levels of 0.00 electron volts. Zer 3.50 electron volts and 5.50 electron volts. That's our final answer. What we're ultimately trying to figure out is we're trying to figure out what the wavelengths are for these three energy levels. Awesome with that in mind, let's look at our multiple choice answers to see what our final answer pair might be. So for a lambda 12 is equal to 434 nanometers and lambda 23 is equal to 248 nanometers B is Lambda 12 is equal to 355 nanometers. And Lambda 23 is equal to 248 nanometers C is Lambda 12 is equal to 434 nanometers. And LAMBDA 23 is equal to 621 nanometers. And finally, for D LAMBDA 12 is equal to 355 nanometers. And LAMBDA 23 is equal to 621 nanometers. Awesome. So first off, let us recall that for each quantum of light, it will have an energy of E equals Planck's constant multiplied by the frequency which is equal to Planck's constant multiplied by the speed of light all divided by the wavelength of light. So note that we can rearrange this energy equation to isolate and solve for lambda. And when we do, we will find that Lambda is equal to H multiplied by C all divided by capital E. So now at this stage, we must note that we have three energy levels. So at N equals one, we have 0.00 electron volts at energy level two. So N equals two, we have 3.50 electron volts and at energy level three. So N equals three, we have 5.50 electron volts. So at this point, we need to determine the absorption energies from N equals one to N equals three. So uh for focusing on e the energy of 1 to 2. So from 1 to 2, and it'll equal 3.5 electron volts. And from energy level 2 to 3, it will equal 5.5 electron volts minus 3.50 electron volts, which is equal to 2.00 electron volts. So now at this stage, we can solve for our final answers for the wavelengths. So with that in mind, let's note that lambda 12, so energy level 1 to 2 is equal to h multiplied by C divided by e the energy levels of 1 to 2, which is equal to. So let's note really quickly here that the numerical value for Planck's constant is 4.14 multiplied by 10 to the power of negative 15. And its units are electron volts multiplied by seconds multiplied by the speed of light, which is 3.0 multiplied by 10 to the power of 8 m per second, divided by 3.50 electron volts. So when we plug that into our calculator, we will find that lambda one tube is equal to 3.55 multiplied by 10 to the power of negative 7 m. But we need to convert meters to nanometers. So you can use dimensional analysis or you can just multiply by 10 to the power of nine, we will find that it is converted to 355 nanometers. And note that when it's in meter units I round it to two decimal places. OK. So following a similar methodology to solve for lamb to 23. So from energy levels 2 to 3 is equal to pons constant multiplied by the speed of light divided by the energy level from 2 to 3 which is equal to. So plugging in our known variables again. So 4.14 multiplied by 10 to the power of negative 15, the electron volts multiplied by seconds multiplied by the speed of light. So 3.0 multiplied by 10 to the power of 8 m per second, all divided by 2.00 electron volts. So when we plug it into a calculator, we will find that Lambda 23 is equal to 6.21 multiplied by 10 to the power of negative 7 m which multiplying it by 10 to the power of nine, we will find that it's equal to drum roll, 621 nanometers. And these are our final answers. Hooray, we did it. So looking at our multiple choice answers and we had to convert meters to nanometers because as you can see all of our multiple choice answers are in nanometers. So looking at our multiple choice answers, the correct answer has to be the letter D Lambda. 12 is equal to 355 nanometers. And Lambda 23 is equal to 621 nanometers. Thank you. So much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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The allowed energies of a simple atom are 0.00 eV, 4.00 eV, and 6.00 eV.c. What wavelengths appear in the atom’s absorption spectrum?

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