What is the value of ni for an electron

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The discussion revolves around calculating the initial energy level (ni) of an electron in a hydrogen atom that emits a photon with a wavelength of 93.14 nm as it transitions to the ground state. Participants emphasize the importance of using the Rydberg formula to relate the wavelengths and energy levels of the electron transitions. The conversion of the wavelength from nanometers to meters is noted, and calculations involving the Rydberg constant are discussed to derive the necessary values. There is a focus on understanding how to manipulate the formula to isolate ni, with some participants sharing their attempts and confusion regarding the calculations. Ultimately, the conversation aims to clarify the steps needed to find the correct initial energy level for the electron's transition.
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Homework Statement



what is the value of ni for an electron that emits a photon of wavelength 93.14 nm when it returns to the ground state in the hydrogen atom

Homework Equations



planck's constant 6.63 x 10^-34

The Attempt at a Solution


no idea [/B]
 
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Hi cindy1234,

You need to provide some attempt before help can be given. Start by looking though your course notes and text to find relevant equations pertaining to hydrogen emission lines, wavelength and energy, and so forth.
 
this is as far as i can get still don't get some of this

change 93.14 nm to meters
93.14 x 1 x 10^-9 = 9.314 x 10^-8

9.314 x 10^-8 = 1.097 x 10^7 rydberg constant

1 - 1 / n exponent 2 subscript 2 = 0.9787 (how did they get this)
 
There's a particular formula that applies here that should be in your notes or text. It's probably in the chapter that introduces quantum theory as it pertains to atoms and light emission/absorption. Hint: the constant that you named above is used in the formula of the same name.
 
chemistry text - has only one example of an electron transition - that is calculate the wavelength in nm of the photon emitted when an electron transitions from the n = 4 state to the n = 2 state in a hydrogen atom - which i can do but i don't know how to change this to get the n subscript i

tried changing to m and 1 / 9.314 x 10^-8 and get 1.074 x 10^7 as an answer

found the following on the internet
wavelength = 430 nm
find n i
(1/n 1 ^2 - 1 / n 2 ^2) = 2325581 / (1.09677581 x 10^7)
(1/2^2 - 1/n2^2) = 0.212038
1/n2^2 = 0.25 - 0.212038
n2^2 = 26.342 (only part i don't understand) (only need help with this part)
n^2 = 5 (square root)
 
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cindy1234 said:
chemistry text - has only one example of an electron transition - that is calculate the wavelength in nm of the photon emitted when an electron transitions from the n = 4 state to the n = 2 state in a hydrogen atom - which i can do but i don't know how to change this to get the n subscript i
Concentrate on finding the value for the ##n_i## rather than its subscript. You know the sequence of integers that the n's are drawn from, so you can later match the value of n to a subscript.
tried changing to m and 1 / 9.314 x 10^-8 and get 1.074 x 10^7 as an answer

found this on internet
change nm to m 93.14 nm to 9.314 x 10^-8 m

1 - 1 / exponent 2 n subscript 2 = 0.9787 (how did they get this)
1/ exponent 2 n subscript 2 = 0.02128 (how did they get this)
n exponent 2 subscript 2 = 46.98 (how did they get this)
n subscript 2 = 6.85 rounded to 7 (square root of 46.98)
They started with the Rydberg formula (which your chemistry text should have). Your problem statement states that the electron transitions to the ground state, giving you one of the n values. The other n value is then the unknown that you're looking for.

Note that you can use the x2 and x2 buttons in the edit window top bar menu to create subscripts and superscripts for your formulas.
 
cindy1234 said:
found the following on the internet
wavelength = 430 nm
find n i
(1/n 1 ^2 - 1 / n 2 ^2) = 2325581 / (1.09677581 x 10^7)
(1/2^2 - 1/n2^2) = 0.212038
1/n2^2 = 0.25 - 0.212038
n2^2 = 26.342 (only part i don't understand) (only need help with this part)
Both sides were inverted:

(1/n2^2)-1 = (0.25 - 0.212038)-1
n2^2 = (0.037962)-1
n2^2 = 26.342
n2 = √26.342
n2 ≈ 5
 
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