A biophysicist grabs the ends of a DNA strand with optical tweezers and stretches it \(26 \mu \mathrm{m}\). How much energy is stored in the stretched molecule if its spring constant is \(0.046 \mathrm{pN} / \mu \mathrm{m} ?\)

Short Answer

Expert verified
The energy stored in the DNA strand when stretched is found by substituting the values of the spring constant and displacement into the formula for elastic potential energy. Upon calculation, the energy comes out to be approximately \( 1.548 \times 10^{-20} \) Joules.

Step by step solution

01

Convert Units

The spring constant given is in \( \mathrm{pN}/\mu \mathrm{m} \) and displacement \( x \) is in \( \mu \mathrm{m} \). But the SI unit of force is Newton (N) and that of displacement is meter (m). So we need to convert \( \mathrm{pN} \) to \( \mathrm{N} \) and \( \mu \mathrm{m} \) to \( \mathrm{m} \) before substituting the values into the formula. \1 piconewton (pN) = \(1 \times 10^{-12} \) Newtons (N) \Therefore, \( k = 0.046 \times 10^{-12} \) N/m \1 micrometer (\( \mu \mathrm{m} \)) = \( 1 \times 10^{-6} \) meters (m) \Therefore, \( x = 26 \times 10^{-6} \) m
02

Calculate energy

Substitute \( k \) and \( x \) into the formula \( U = \frac{1}{2} k x^2 \). \The elastic potential energy stored in the DNA strand will be \\( U = 0.5 \times (0.046 \times 10^{-12} \) N/m\() \times (26 \times 10^{-6} \) m\()^2 \)

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