Calculating the Effective Spring Constant of a Charged DNA Molecule

AI Thread Summary
The discussion revolves around calculating the effective spring constant of a charged DNA molecule, which is 2.09 µm long and compresses by 1.09% when ionized. One participant calculated the force (F) as 5.27 x 10^-17 N and the displacement (x) as 2.27 x 10^-8 m, leading to a spring constant (k) of 2.31 x 10^-9 N/m. However, another participant suggests that the force value seems too low and requests clarification on the calculations used to derive it. They agree on a similar distance between charges of approximately 2.06 x 10^-6 m. The conversation emphasizes the importance of accurate calculations in determining the effective spring constant.
Kdapik
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A molecule of DNA (deoxyribonucleic acid) is 2.09 µm long. The ends of the molecule become singly ionized: negative on one end, positive on the other. The helical molecule acts like a spring and compresses 1.09% upon becoming charged, and remains in this equilibrium position. Determine the effective spring constant of the molecule. I did calculate F and I got 5.27*10^-17 then I found x which is 2.27*10^-8
I found k from the equation k= f/x
I got 2.31*10^-9 N/m
But still, it's the wrong answer :(
 
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Hello Kdapik. Welcome to PF!

I get a somewhat larger value for the force F. What did you use for the distance between the charges when you calculated F?
 
TSny said:
Hello Kdapik. Welcome to PF!

I get a somewhat larger value for the force F. What did you use for the distance between the charges when you calculated F?
2.06*10^-6 m
 
OK, that's pretty close to what I get (2.07 x 10-6 m). I still don't see how you are getting 5.27 x 10-17 N for the force. Can you show all the numbers that you used to get F?
 

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