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astra-53 [7]
3 years ago
13

What's the difference between inplicit and explicit solvent?

Physics
1 answer:
marusya05 [52]3 years ago
7 0
Hi 

<span>Explicit- and implicit-solvent molecular dynamics simulations are performed to study complexation in two polyelectrolyte systems: poly(styrene sulfonate)/poly(allylamine hydrochloride) (PSS/PAH) and poly(acrylic acid)/poly(allylamine hydrochloride) (PAA/PAH). Both these systems have been used in layer-by-layer assembly of polyelectrolyte films, with the first of these typically yielding linear growth in film thickness with increasing numbers of layers and the second yielding exponential growth. In both the systems, the polyelectrolytes have the same number of monomers and are present in stoichiometric proportion, and water is used as solvent. Simulations give important insights into the structure and composition of the complexes. We found that the PSS/PAH complex is more compact and has a smaller fraction of water than the PAA/PAH complex due to the presence of phenyl rings in the PSS chain and the hydrophilic nature of the charged group in the PAA chain. Either an increase in salt concentration or a decrease in partial charge fraction increases swelling and the water content in the complex. To overcome the computational limitations associated with the explicit-solvent simulation, a simple strategy to develop the force field for an implicit-solvent simulation is proposed. In the absence of salt and for fully charged polyelectrolytes, the radius-of-gyration and the various radial distribution functions predicted by the implicit-solvent simulation match well with those predicted by the explicit-solvent simulation, while reasonable agreement is obtained in the other cases. The implicit-solvent simulation was performed for bigger system sizes, and we observed trends similar to those observed for smaller system sizes, suggesting that the simulation results are independent of system size.
</span>
hope it helps..
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Answer:

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Mandarinka [93]

Answer:

145 m

Explanation:

Given:

Wavelength (λ) = 2.9 m  

we know,

c = f × λ  

where,

c = speed of light ; 3.0 x 10⁸ m/s

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thus,

f=\frac{c}{\lambda}

substituting the values in the equation we get,

f=\frac{3.0\times 10^8 m/s}{2.9m}

f = 1.03 x 10⁸Hz  

Now,

The time period (T) = \frac{1}{f}

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the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s  

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