Ionic Array
Posted in Natural Care by admin
Physics: Electrical Energy and Potential?
Hi, I have some physics questions, perhaps you could check my answers?
1. A KBr ionic molecule consist of a K+ ion and a Br- ion separated by an equilibrium distance of 0.33 nm. What is the magnitude of the Coulomb potential energy of the molecule in kcal/mole?
**100.4 kcal/mole
2. A potential difference of 82 mV exists across a cell membrane. The thickness of the membrane is 9.5 nm. What is the magnitude of the electric field inside the membrane (in MV/m)?
**8.63 MV/m
3. What is the magnitude of the voltage change (in mV) across a 18-nm thick membrane of a cell if the electric field in the membrane is 5×10^6 V/m?
**90mV
4. Suppose that a biological membrane has a surface charge density σ = 0.8×10^-4 C/m2. The charge is uniformly distributed over the surface and the average distance between charges is “L”. That is, there is the equivalent of one charge unit e = 1.6×10^-19 every “L” nm in a square array on the surface. Calculate the distance “L” in nm.
**44.7 nm
1. I get PE (for the two ions only) = kQ1Q2/(Kd*R) = 6.99115E-19 J. (k = 1/(4pi*e0) = 8987551787.38888 N-m/C^2.)
For answer, convert J to kcal (mult. by 2.390057e-4) and multiply by Avogadro’s number (6.0221415E23). I get 100.6255 kcal/mole.
2. Agree (MV being megavolts).
3. Agree.
4. Agree.
SCD(e) = 1/(L*1E-9)^2 e/m^2
SCD(e) * 1.6E-19 C/e = 8E-5 C/m^2
1/(L*1E-9)^2 * 1.6E-19 = 8E-5 ==>
(L*1E-9)^2 = 1.6E-19/8E-5 ==>
L = sqrt(1E18*1.6E-19/8E-5) = 44.72136 nm
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