Balmer-Rydberg Equation Questions (1 Viewer)

[scardizzle]

I've been revising quanta to quarks for my trials, problem is our teacher never taught it to us -_-. So forgive me if my questions are stupid.

First question comes from jacaranda q3 on p435

The radius of the orbit of an electron in the ground state of the hydrogen atom is 5.3 x 10^-11m. Calculate the radius of an electron when it is in each of the following states:
a) n = 2
b) n = 3
c) n = 4

And one from dot point:
2 lines in the spectrum of an atom have frequencies of 4.0 x 10^14 and 5.6 x 10^14 Hz

a) calculate the wavelength of the photons responsible for these spectral lines
b) Predict the frequency of a 3rd spectral line
c) Account for your answer b)

Any help would be very appreciated.

 

[shady145]

i dnt think ive done this yet but for 2.a) couldnt u use v=fxwavelength
wavelength=velocity/frequency
v=3x10^8
so wavelength=3x10^8/4.0 x 10^14

 

[Aerath]

I can't do the Jacaranda question. =\

But the dotpoint question:
a)Just use v = f*wavelength. Answers in the textbook are wrong, it should be 750nm and 536nm.
b) Based on the assumption that the first value is a level 2 to 1 transition, and the next is a level 3 to 1 transition. They're asking for what a level 3 to 2 transition would be. Given that, the difference would just be 5.6x 10^14 - 4x10^14 = 1.6x10^14 Hz
c) Explanation in b).

 

[scardizzle]

I can't do the Jacaranda question. =\

But the dotpoint question:
a)Just use v = f*wavelength. Answers in the textbook are wrong, it should be 750nm and 536nm.
b) Based on the assumption that the first value is a level 2 to 1 transition, and the next is a level 3 to 1 transition. They're asking for what a level 3 to 2 transition would be. Given that, the difference would just be 5.6x 10^14 - 4x10^14 = 1.6x10^14 Hz
c) Explanation in b).

Okay that clears a) up but with b) how did you know that the energy values were from the 2nd to the the 1st shell and 3rd to 1st shell?

Also with the jacaranda maybe the circumference is the wavelength? Since a period is one revolution and the distance for one revolution is the circumference? Not sure if that helps

 

[darkchild69]

to find the radius its simply the following:

n^2*r

so for n=2 the radius would be 4x, n=3 would be 9x and n=4 would be 16x. You are not required to know this, nor the derivation.

Aerath, why did you use the initial energy level as n=1 in those questions? The HSC syllabus is only concerned with the Balmer series, which has an initial energy level n=2. n=1 corresponds to the lyman series (UV light) which should not need to be covered iirc.

That's a pretty piss poor question if you ask me, but it seems like you went the right way about it.

 

[Aerath]

Aerath, why did you use the initial energy level as n=1 in those questions? The HSC syllabus is only concerned with the Balmer series, which has an initial energy level n=2. n=1 corresponds to the lyman series (UV light) which should not need to be covered iirc.

Hmmm, I distinctly remember that for that particular question in Shadwick's Physics Dotpoint book, they asked us to talk about the Lyman and Balmer series, so I assumed it was ok to use n = 1.

 

[darkchild69]

Hmmm, I distinctly remember that for that particular question in Shadwick's Physics Dotpoint book, they asked us to talk about the Lyman and Balmer series, so I assumed it was ok to use n = 1.

Looking at the syllabus there are two main points:

-Process and present diagrammatic information to illustrate Bohrs explanation of the Balmer series

- Solve problems and analyse information using the Balmer-Rydberg equation

So i actually suppose you could be assessed on different initial energy states rather than n=2, but they wouldnt be able to use the term Lyman. They may say between n=3 and n=1, you are right!

my bad

 

[Bernie Mac]

Did Jordan start this post? Cos it seems like it.

 

[scardizzle]

Did Jordan start this post? Cos it seems like it.

I think you got me confused with someone else.........

 

[cutemouse]

problem is our teacher never taught it to us -_-.

Hahah how's Ball treating you guys? XD

 

[scardizzle]

Hahah how's Ball treating you guys? XD

whoa, how did you work that one out?

EDIT: I see what you did there jason minh very clever

 

[cutemouse]

lol which other teacher in the state claims that you can't find the gradient of a curved graph?

 

[Mc Fadge]

I guess a logical way of working out the Jacaranda question is that since electrons exist in orbits of stable wavelengths, the circumference of the radius is the wavelength. Therefore, the next orbit will occur at 2x the inital wavelength as it will produce the next standing wave.