Question

An accelerator produces a beam of protons with a circular cross section that is 2.0 mm in diameter and has a current of 1.0 mA. The current density is uniformly distributed through the beam. The kinetic energy of each proton is 20 MeV. The beam strikes a metal target and is absorbed by the target. (a) What is the number density of the protons in the beam

Answer 1

the number density of the protons in the beam is 3.2 × 10¹³ m⁻³

Step-by-step explanation:

Given that;

diameter D = 2.0 mm

current I = 1.0 mA

K.E of each proton is 20 MeV

the number density of the protons in the beam = ?

Now, we make use of the relation between current and drift velocity

I = MeAv ⇒ 1 / eAv

The kinetic energy of protons is given by;

K =
`(1)/(2)`
`m_(p)`v²

v = √( 2K /
`m_(p)` )

lets relate the cross-sectional area A of the beam to its diameter D;

A =
`(1)/(4)`πD²

now, we substitute for v and A

n = I /
`(1)/(4)`πeD² ×√( 2K /
`m_(p)` )

n = 4I/π eD² × √(
`m_(p)` / 2K )

so we plug in our values;

n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )

n = 1.98695 × 10¹⁸ × 1.6157967 × 10⁻⁵

n = 3.2 × 10¹³ m⁻³

Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³