Calculate Wavelength from Rydberg's Constant

[CDerhammer]

Wavelength HELP!

Homework Statement

Calculate the wavenumber (v=1/lambda) and the wavelength of the first transition in the visible region of the atomic spectrum of hydrogen.
I KNOW THE ANSWER... 656.3 nm. I just don't have a clue how they got that answer, I do know that I had to use Rydberg's Constant (1.097X10^7 m ^-1) but I have no idea what to do or how to do it. Thank you anyone who wants to take time and help me. And I don't know if this will mean anything but it is Inorganic Chem 430.

Homework Equations

1/λ=R(1/n₁²-1/n₂²)

The Attempt at a Solution

I have been at this for hours trying to find out how to calculate everything. and IDK if it's my ti-84 plus edition that is doing the math wrong or what but I cannot find out how to compute this...

Homework Statement

No variables were given to me except rydberg's constant.

 

[netgypsy]

If you Google the first transition for hydrogen you get from 3 to 2 for your n values.

From there just put the numbers into the equation, evaluate the right side and take the reciprocal to get the wavelength.

 

[sandy.bridge]

Look at the atomic spectrum for hydrogen. There are only four lines. Which values of "n" will you be using for the first transition?

 

[CDerhammer]

If you Google the first transition for hydrogen you get from 3 to 2 for your n values.

From there just put the numbers into the equation, evaluate the right side and take the reciprocal to get the wavelength.

Okay, Now i get 1/λ=1.097x10⁷m^-1(1/6) am I on the right track?

 

[CDerhammer]

Look at the atomic spectrum for hydrogen. There are only four lines. Which values of "n" will you be using for the first transition?

my book says it starts at n=2 and n₂≥3

 

[netgypsy]

1/2squared - 1/3squared does not equal 1/6 . You'll get 1/4 - 1/9 =5/36, not 6/36 otherwise you're doing it right. You just had to figure out what the n values were for that first transition which as I said, I asked Google and it said from 3 to 2.

 

[CDerhammer]

1/2squared - 1/3squared does not equal 1/6 . You'll get 1/4 - 1/9 =5/36, not 6/36 otherwise you're doing it right. You just had to figure out what the n values were for that first transition which as I said, I asked Google and it said from 3 to 2.

Whoops yeah, I meant 5/36. then I get 1/λ=1523611.111m^-1
lol, sorry if I am bugging you but what do I do now to convert it to nm? Particularly 656.3 nm... lol

 

[netgypsy]

If you google nanometer you will see that 1 nm is 10 ^ -9 meters :-)

You found 1/lamda You need to take the reciprocal to get lamda, the wavelength. Than change it to nanometers.

 

[CDerhammer]

If you google nanometer you will see that 1 nm is 10 ^ -9 meters :-)

You found 1/lamda You need to take the reciprocal to get lamda, the wavelength. Than change it to nanometers.

Hmm... I am not too sure how to get the number 656.3 though. I am fine with converting. Just can't find the 656.3

 

[netgypsy]

When you solve the equation you'll get 1/lamda = 5/36 (1.097 X l0^ 7)

flipping the whole thing over you'll get lamda = 36/5divided by 1.097 X 10 ^7

which is pretty close to 7 divided by 1.1 X 10 ^ 7 which you can see by estimating is 6 something X 10 -7 it will come out the right answer.

 

[CDerhammer]

Wow. The book says nothing about flipping everything. got the answer perfectly. Thanks!
*I MUST REMEMBER TO FLIP EVERYTHING!

 

[netgypsy]

There are quite a number of equations in physics that allow you to find the reciprocal of quantities and whenever the thing you are looking for is in the denominator like with 1/lamda, you have to calculate the number part and then take the reciprocal to get the thing you want, lamda. And one of the easiest ways to do that is to just flip the whole thing over on both sides (but be careful because you can NOT flip the sum of two fractions. You have to first get a common denominator, add, then flip.

 

[CDerhammer]

There are quite a number of equations in physics that allow you to find the reciprocal of quantities and whenever the thing you are looking for is in the denominator like with 1/lamda, you have to calculate the number part and then take the reciprocal to get the thing you want, lamda.

Well this is going to be a long journey through my inorganic chem 430 class lol. How come I couldn't find this stuff on youtube or google or anything?

 

[netgypsy]

It's on there. I found the n values on google. You would get that same problem in physics, in several different areas. I never took inorganic. LOVED Physical chem and instrumental analysis.

 

[CDerhammer]

It's on there. I found the n values on google. You would get that same problem in physics, in several different areas. I never took inorganic. LOVED Physical chem and instrumental analysis.

Yeah I'm a biochem major at Indiana University, I have physical chem 362 coming in the spring. but I am only a freshman lol

 

[netgypsy]

Good luck and enjoy. I could have stayed in college forever. Sooo much fun. so many interesting people.