Question

A molecule of DNA (deoxyribonucleic acid) is 2.26 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 0.99% upon becoming charged. Determine the effective spring constant of the molecule

Answer 1

Given that,

L = 2.26*10^(-6) m

compression in spring, L = 0.99 %

L = (0.99 / 100)* 2.26*10^(-6) = 0.0223*10^(-6) m

Now, length of DNA molecule,

L' = L - L

L' = 2.2376*10^(-6) m

Here,

spring force = electrostatic force

K* L = kq^2 / L'^2

(q = e = 1.6*10^(-19))

K*0.0223*10^(-6) = 9*10^9*(1.6*10^(-19))^2 /  (2.2376*10^(-6))^2

K = 2.05*10^(-9) N/m

Hence, effective spring constant of the molecule ,

K = 2.05*10^(-9) N/m