## Atomic Spectra

H-Atom ($E_{n}=-\frac{hR}{n^{2}}$)

Archana Biju 1G
Posts: 29
Joined: Fri Sep 28, 2018 12:18 am

### Atomic Spectra

Can someone explain why the transition from n = 4 to n = 2 emits radiation of longer wavelength than the transition from n = 5 to n = 1?
Thank you!

julia_lok_2K
Posts: 56
Joined: Fri Sep 28, 2018 12:25 am

### Re: Atomic Spectra

I think this is because a smaller amount of energy is emitted when an electron goes from n=4 to n=2 compared to going from n=5 to n=1 since the energy difference is smaller. This energy can be represented by E=hv, from which the equation E=(ch)/wavelength, or wavelength=(ch)/E, can be derived. From the latter equation, we can see that as E decreases, the wavelength increases. Therefore, an electron going from n=4 to n=2 will emit radiation of a longer wavelength then that of an electron going from n=5 to n=1.

Hope this helps!

Ethan Yi 1K
Posts: 62
Joined: Fri Sep 28, 2018 12:28 am

### Re: Atomic Spectra

Just to reiterate, when the energy difference is smaller, wavelength is larger, and frequency is lower.

Sophia Fox 4B
Posts: 29
Joined: Fri Sep 28, 2018 12:27 am

### Re: Atomic Spectra

Since energy is directly proportional to frequency, its inversely related to wavelength (c = wavelength x freq and E = planks constant (h) x freq ==> derive (E = hc/wavelength) by plugging in for freq). There is a greater change in energy from n =5 to n = 1 than n = 4 to n = 2, so the transition from 5 to 1 gives off a smaller wavelength than the one from 4 to 2. Does that make sense? It's really just thinking about the equations and how the variables relate to each other.

Bingcui Guo
Posts: 30
Joined: Fri Sep 28, 2018 12:19 am

### Re: Atomic Spectra

You can use the formula to calculate the different energy and thus compare. Concrete number is the answer. Hope this will help.

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