CBSE Class 12 Physics 2022 Term 2 Outside Delhi Set 1 Solved Paper

Question : 6

Total: 12

A narrow beam of protons, each having 4.1‌MeV energy is approaching a sheet of lead (Z=82). Calculate:
(i) the speed of a proton in the beam, and
(ii) the distance of its closest approach. 3

Solution:

(i) KE of 4.1MeV proton
=4.1×106×1.6×10−19J
Mass of proton =1.67×10−27kg
v2=

2KE

Or, v2=

2×4.1×106×1.6×10−19

1.67×10−27

Or, v2=7.85×1014
∴v=2.8×107m∕s
(ii) Energy of proton =4.1MeV
Atomic number(Z) of lead =82
When the proton is at distance of closest approach (r), then
KE of the system =0
PE of the system =

kZe2

So, from the conservation of energy principle,
4.1MeV=0+

kZe2

Or, r0=

kZe2

(4.1×106e)

Or, r0=

9×109×82×e2

4.1×106

m
∴r0=288×10−16m
r0=2.88×10−14

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