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

# A photon is emitted when a hydrogen atom undergoes a transition from the $n=5$ state to the $n=3$ state. Calculate (a) the energy (in electron volts), (b) the wavelength, and(c) the frequency of the emitted photon.

## a) 0.968 $\mathrm{eV}$b) 1.28$\mu \mathrm{m}$c) $2.34 \times 10^{14} \mathrm{Hz}$

Quantum Physics

Atomic Physics

### Discussion

You must be signed in to discuss.
##### Christina K.

Rutgers, The State University of New Jersey

##### Andy C.

University of Michigan - Ann Arbor

LB

### Video Transcript

for part A to calculate the energy of a photon that is released from an Adam during the transition between the in equals five to the n equals three state. We're gonna use the equation that says the energy of that photons e is equal to negative. 13.6 electron volts were negative. 13.6 electron volts comes from the ground state energy of the hydrogen atom. We're dealing with a hydrogen atom here. So and then this is multiplied by one over the initial state squared, which is five squared minus one over the final state squared, which is three. So this is 1/3 square. So if we carry out this operation, we find that the energy of this photon and electron volts is equal to zero 0.968 electron volts for part B. It wants us to use this information to calculate the wavelength. So to calculate the wavelength, we're going to use the, uh, the fact that energy is related to wavelength through the equation. E is equal to thanks constant h times the speed of light see divided by Linda the wavelength. So we can rearrange this equation to solve for Lambda we have. Lambda is equal to h time. See over the energy. Since our energy is in units of electron volts, we want to use a value of plank's constant that has units of electron volt seconds instead of Jewell seconds. So an electron volt seconds plank's constant is 4.1357 times 10 to the minus 15 electron volt seconds. The speed of light is still in meters per second, so it's three times 10 to the eight meters per second. Carrying out this calculation, we find that the wavelength is equal to 1.28 times 10 to the minus six meters or 1.28 micro meters. You can write it either way. I'm gonna write it as micro meters. But you could also write it as 10 to the minus six meters. Then lastly for part C were asked to find the frequency. And this can easily be done from the wavelength. So the frequency by definition is equal to the speed at which the object is moving. In this case, it's light. So it's see, divided by the wavelength of that object, plugging these values and to make sure you convert wavelength back two meters so it's 1.28 times 10 to the minus six meters. Must be delight is in meters per 2nd 3 times 10 to the eight meters percent. This comes out to equal to 0.23 times 10 to the 14 hurts were hurt is the same as a unit of one over seconds. But we're gonna use units of hurts since that is the common S I unit for frequency. Welcome box. It is our solution for part C.

University of Kansas

#### Topics

Quantum Physics

Atomic Physics

##### Christina K.

Rutgers, The State University of New Jersey

##### Andy C.

University of Michigan - Ann Arbor

LB