Question

A narrow beam of protons, each having 4.1 MeV energy is approaching a sheet of lead (Z = 82). Calculate:

1. the speed of a proton in the beam, and
2. the distance of its closest approach

Answer 1

(i) KE of 4.1 MeV proton = 4.1 × 10⁶ × 1.6 × 10⁻¹⁹ J

Mass of proton = 1.67 × 10⁻²⁷ kg

v² = `(2"KE")/"m"`

Or, v² = `(2 xx 4.1 xx 10^6 xx 1.6 xx 10^-19)/(1.67 xx 10^-27)`

Or, v² = 7.85 × 10¹⁴

∴ v = 2.8 × 10⁷ MeV

(ii) Energy of proton = 4.1 MeV

Atomic number (Z) of lead = 82

When the proton is at the distance of the closest approach (r), then,

KE of the system = 0

PE of the system = `"kZe"^2/"r"`

So, from the conservation of energy principle,

4.1 MeV = `0 + "kZe"^2/"r"`

O, r₀ = `"kZe"^2/((4.1 xx 10^6  "e"))`

Or, r₀ = `(9 xx 10^9 xx 82 xx "e"^2)/(4.1 xx 10^6"e")"m"`

∴ r₀ = 288 × 10⁻¹⁶ m

r₀ = 2.88 × 10⁻¹⁴