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प्रश्न
A narrow beam of protons, each having 4.1 MeV energy is approaching a sheet of lead (Z = 82). Calculate:
- the speed of a proton in the beam, and
- the distance of its closest approach
उत्तर
(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
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