A Van de Graaff generator produces a beam of 2.02-MeV deuterons, which are heavy hydrogen nuclei containing a proton and a neutron. A) If the beam current is 10.0 μA, how far apart are the deuterons?
B) Is the electrical force of repulsion among them a significant factor in beam stability? Explain.

A)  The distance of the deuterons from one another  = 2.224× 10⁻⁷ m

B)  The electrical force of repulsion among them shows a small effect  in beam stability.

Explanation:

Given that:

A Van de Graaff generator produces a beam of 2.02-MeV deuterons

If the beam current is 10.0 μA, the distance of the deuterons from one another can be determined by using the concept of kinetic energy of the generator.

2 K.E = mv²

so, v is the velocity of the deuterons showing the distance of the deuterons apart from one another.

v = 13911611.49  m/s

v = 1.39 × 10⁷ m/s

So, If the beam current is 10.0 μA.

We all know that:

t = 1.6 × 10⁻¹⁴ s

Finally, the distance of the deuterons from one another  = v × t

the distance of the deuterons from one another  = (1.39 × 10⁷ m/s × 1.6 × 10⁻¹⁴ s)

the distance of the deuterons from one another  = 2.224× 10⁻⁷ m

B) Is the electrical force of repulsion among them a significant factor in beam stability? Explain.

The electrical force of repulsion among them shows a small effect  in beam stability. This is because, one nucleus tends to put its nearest neighbor at potential V = (k.E × q) / r = 7.3e⁻⁰³ V. This is very small compared to the 2.02-MeV accelerating potential, Thus, repulsion within the beam is a small effect.

we need answer choices

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.