Amolecule of dna (deoxyribonucleic acid) is 2.23 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 1.13% upon becoming charged. determine the effective spring constant of the molecule.

End of the molecule become singly ionized mean charge on both the sides 
Q = 1.6 * 10 ^ -19 C

F = KQ^2 / r^2 
F = ( 8.99 * 10 ^-9 ) (1.6 * 10 ^-19) ^2 / (2.23 * 10 ^-6) ^2 
F =  4.627 * 10^-17  N

F= Kx
x = 0.0113 ( 2.23 * 10^-6)

put values in F = Kx 

4.627 * 10 ^-17  =   K  ( 0.0113 ( 2.23* 10 ^-6) 
K =   4.627  * 10 ^ -17  /   2.519 * 10 ^ -8 
K =  1.83c 10 ^-9 N/m

Wish i can but don’t know this