Atomic Spectra Post-Assesment

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SamanyaCMuyunga1K
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Atomic Spectra Post-Assesment

Postby SamanyaCMuyunga1K » Sat Oct 31, 2020 4:39 pm

43. In the hydrogen atomic spectrum, what is the wavelength of light associated with the n = 2 to n = 1 electron transition?

I tried to find the wavelength by using the equation v= R [1/(n1)^2 - 1/(n2)^2], and then plugging in the value for v in wavelength = c/v, but the answer I came up with was not in any of the answer choices. How do you solve for this problem?

Mackenzie Fernandez 3G
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Joined: Wed Sep 30, 2020 10:07 pm
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Re: Atomic Spectra Post-Assesment

Postby Mackenzie Fernandez 3G » Sat Oct 31, 2020 4:43 pm

SamanyaCMuyunga1K wrote:43. In the hydrogen atomic spectrum, what is the wavelength of light associated with the n = 2 to n = 1 electron transition?

I tried to find the wavelength by using the equation v= R [1/(n1)^2 - 1/(n2)^2], and then plugging in the value for v in wavelength = c/v, but the answer I came up with was not in any of the answer choices. How do you solve for this problem?


hi!
I found it best for these equations to not use v but (1/). Then use R=1.0974x10^7 (since it involves Hydrogen).
Hope this helps!

Truman Chong Dis 3G
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Re: Atomic Spectra Post-Assesment

Postby Truman Chong Dis 3G » Sat Oct 31, 2020 4:55 pm

I’m not quite sure what was your work for this problem but here’s what I did. For the Rydberg formula, I noticed that n1 should always be the smaller energy level number (in this case 1); therefore, n2 would be 2. Plugging in the values, squaring them, and then subtract them. After multiplying by R I got 2.4673...x10^15. Plugging this frequency into ν= c/λ, I got v to equal 1.215...x10^-7 which is answer choice D.

Bailey Giovanoli 1L
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Joined: Wed Sep 30, 2020 9:50 pm

Re: Atomic Spectra Post-Assesment

Postby Bailey Giovanoli 1L » Sat Oct 31, 2020 4:57 pm

I'm not sure if this is an issue for you, but I always forget to take the square root to find the n instead of n^2. So, maybe check if that's an issue. If that's not the issue, I would be willing to help you further.


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