How do you find n2?
The original problem asks us to determine n value for initial and final energy levels of the electron when a line is observed at 102.6nm. I first started by using v=c/λ and got 2.922x10^15 s^-1. Then I used 1/n2^2 = 1/n1^2 - v/R to find the second level, but I don't think I understand how to use the equation, I got stuck when I got 1/n2^2=0.112. In the book, I saw that the answer is 3 but I don't understand how they got that.
Thank you!
UV Spectrum of H-Atom
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Re: UV Spectrum of H-Atom
Once you get to 1/n2^2=0.112 you can just do it as an algebra problem!
Multiply both sides by n2^2
1=0.112*n2^2
Divide both sides by 0.112
8.93=n2^2
Square root of both sides
2.988=n2
Round to the nearest whole number because n values are always integers --> then you'll get n2=3
Multiply both sides by n2^2
1=0.112*n2^2
Divide both sides by 0.112
8.93=n2^2
Square root of both sides
2.988=n2
Round to the nearest whole number because n values are always integers --> then you'll get n2=3
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