Our Discord hit 10K members! 🎉 Meet students and ask top educators your questions.Join Here!

Like

Report

Numerade Educator

Like

Report

Problem 21 Medium Difficulty

(a) If an electron makes a transition from the $n=4$ Bohr orbit to the $n=2$ orbit, determine the wavelength of the photon created in the process. (b) Assuming that the atom was initially at rest, determine the recoil speed of the hydrogen atom when this photon is emitted.

Answer

a) 486 $\mathrm{nm}$
b) 0.817 $\mathrm{m} / \mathrm{s}$

Discussion

You must be signed in to discuss.
Top Physics 103 Educators
Elyse G.

Cornell University

Andy C.

University of Michigan - Ann Arbor

Jared E.

University of Winnipeg

Meghan M.

McMaster University

Video Transcript

in this exercise. We have an election transitions from the initial level and I equals two for the fourth energy level of the hydrogen atom to the final. Had the final energy level enough to the second enter your level. And in the first question, we have to calculate what is the wavelength of the folding emitted. So we know that the wavelength of the full term actually one over the wavelength is given by the Redbird constant R H times one over, and I square the initial one over the initial interview level. Square mine is one over the final interview level square. Okay, so Lunda is one over our age. One over times went over in the and I squared minus one over an F square. Rh is 1.97 times 10 to the seven meters to the minus. One times one over and I square. So that's four square 16 that is one over and 1/2 square, just two square. So it's four. So Lunda is equal to 4.86 times 10 to the minus seven meters or 486 centimeters. Okay, this is the answer to question a and in question. Do you have to calculate the recoil speed of the hydrogen atom? So notice that initially the hydrogen atom is breast. So the initial momentum, the total mission momentum of the system is zero and using conservation of momentum the final, the total final Ah, momentum of the system will also be zero in the final woman. Some other system consists of the momentum off the photo P gamma, plus the momentum of the hydrogen atom pH Okay, so I have a big P gamma equals two people minus minus minus pH. And for the reason you have a the magnitude of Big Umma, it was the magnitude of Ph. So I have to do. To complete the speed is to remember that pH The momentum of the the hydrogen atom is the mass off the hydrogen atom, which basically is the mass of the protein. So I'm gonna right and p for a mass of the proton times the speed. Okay, so the speed V is ah, the momentum other falter, which I'm just gonna call p gonna Kim. I'm leaving out the magnitude side over the mass of the proton, M p and we know that P. Gamma the momentum other Fulton is h over Lunda. And this we're gonna multiply by Pi gamma. So this is 6.63 times 10 to the minus 34 jewels. Second, Lunda is for waned 86 times sent to the minus seven meters. And the mass of the protein is given by 1.67 So I'm sin that minus 27 kilograms. So the speed V is equal to yeah 0.817 meters per second. Okay, this is the recall your speed on the hydrogen atoms.

Universidade de Sao Paulo
Top Physics 103 Educators
Elyse G.

Cornell University

Andy C.

University of Michigan - Ann Arbor

Jared E.

University of Winnipeg

Meghan M.

McMaster University