A nanoparticle particle (smaller particle) of mass 1kD (1 Dalton is 1.66 times 1

Question

A nanoparticle particle (smaller particle) of mass 1kD (1 Dalton is 1.66 times 10^-27 kg) is moving along a linear path under a constant force of 3 pN (Fig. 2). Assume that the current position of the particle is 8 nm from the source of influence 2 ps after it starts from the origin. Using Verlet's finite difference algorithm, determine (i.e., complete the following table for cells with question mark): (1) The acceleration of the particle at t = 2 ps (2) The position of the particle at t = 4 ps (3) The velocity of the particle at t = 2s x: position of the particle from a reference point (origin). v: velocity of the particle. a: acceleration of the particle.

Explanation / Answer

Solution :-

According to verlet's finite difference;

xi+1 = xi + (xi - xi-1) + a * dt * dt

In simple words we can use velocity = distance/time

Case 1

Time t = 0

Distance x = 1.10 nm

So velocity = distance/time = 1.10 nm for 0 sec

Acceleration = velocity/time = 1.10 nm/0sec / sec

Case 2:-

Time t = 2 ps

Distance x = 1.22 nm

So velocity = distance/time = 1.22/2 = 0.61nm/s

acceleration = velocity/time = 0.61/2 = 0.305 nm/s^2

Case 3

Time t = 4 ps

Distance x = 8 nm

So velocity = distance/time = 8/2 = 4 nm/s

acceleration = velocity/time = 4/2 = 2 nm/s^2

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