The absorption spectrum of an atom consists of the wavelengths...

Question

The absorption spectrum of an atom consists of the wavelengths 200 nm, 300 nm, and 500 nm. b. What wavelengths are seen in the atom’s emission spectrum?

Accepted Answer

Hey, everyone. So this problem tells us to consider an atom that absorbs radiation at wavelengths of 165 nanometers, 320 nanometers and 510 nanometers from the ground state. We are asked what other wavelengths can be emitted. And we're told to take the ground state energy to be zero electron volts. Our multiple choice answers in units of nanometers are a 338 and 564 B 338 564 and 787 C 243 and 500 or sorry. And 857 or D 243 338 and 857. So the key to solving this problem is first drawing out a energy level diagram. And so we'll have our energy on our y axis and then each energy level on our x axis. And so we're told that our ground state energy or E one is zero electron volts and then we have three other states. So we have an E two E three and E four. And we're told the wavelengths for the transitions from E 23 and 42, E one, that ground state. And so sorry, so we know that there is a wavelength for the transition from E four to E one, from E three to E one and from E two to E one. So what are the other possible transitions here? So we could go from E four to E three, we can go from E 4 to 2 or we could go from E three to E two. And so that finding those wavelengths are going to the other photons that can be emitted. In this situation, you can also recall that the shorter the wavelength, the larger the transition. And that tells us that the transition from E one to E four has the 165 nanometer wavelength from E one T three, we have that 320 nanometer wavelength. And then from E one E two, that is the 510 nanometer wavelength. So to solve this problem, we can recall that our delta E or change in energy is equal to H multiplied by C divided by lambda where H is planks constant C is the speed of light and lambda is our wavelength. And so for each of these energy levels, we can solve for what those energy levels are. And then we can find the delta for each of these transitions. So delta E from 4 to 1 is going to be E four minus E one. And then we know that E one goes to zero. And so E four is equal to that HC divided by Lambda. So we have 6.63 times 10 to the negative 34 dual seconds, four hour planks. Constant speed of light we can recall is three times 10 to the 8 m per second. And then that Lambda from 1 to 4 is 165 nanometers to keep everything in standard units, we will rewrite that as 10 to the negative 9 m. And so that gives us a delta E sorry, just E four of 1.21 times 10 to the negative 18 jos we follow the same process to find E three and E two. So for E three, everything stays the same except for that wavelength which was given as 320 nanometers. And that comes out to 6.22 times 10 to the negative 19 Joel. And similarly, for E two, we have a wavelength of 510 nanometers which gives us an energy of 3.9 times 10 to the negative 19 joules. And so now we can calculate our wavelengths given those transitions. So rearranging this equation, we have LAMBDA is equal to H multiplied by C divided by delta E. And so recalling those transitions that we identified from our energy level diagram, we have Lambda from 42 three. So H again, six times 60 sorry, 6.63 times 10 to the negative 34 D seconds multiplied by our speed of light 10, 3 times 10 to the 8 m per second. All divided by that delta E. So delta E from 4 to 3 will be 1.21 times 10 to the negative 18 joules minus E 36.22 times 10 to the negative 19 joules. We plug that in and we get 3.38 times 10 to the negative 7 m or 338 nanometers. And that is one of the possible wavelengths that can be emitted in this scenario. Similarly, our wavelength from 4 to 2, everything were on the numerator will remain the same. And then we're looking at E four which again is 1.21 times 10 to the negative 18 minus E two, which in this case is 3.9 times 10 to the negative 19. And that comes out to 2.43 times 10 to the negative 7 m or 243 nanometers. Lastly, we have that transition from E three to E two where E three is 6.22 times 10 to the negative 19 jewels. And that comes out to 8.57 times 10 to the negative 7 m or 857 nanometers. And so your possible wavelengths. In addition to those transitions from the ground state are 243 nanometers, 338 nanometers and 857 nanometers. So that's all we have for this one, we'll see you in the next video.

The absorption spectrum of an atom consists of the wavelengths 200 nm, 300 nm, and 500 nm. b. What wavelengths are seen in the atom’s emission spectrum?

Key Concepts

Absorption Spectrum

Emission Spectrum

Energy Level Transitions

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