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CBSE Class 12 · 2024 · Modern PhysicsShort Answer

A proton of energy 1.6 MeV approaches a gold nucleus

(Z = 79)

. Find the distance of its closest approach.

Answer: $r_{0} \approx 7.1 \times 1 0^{- 14} m$ .

Solution

At closest approach all the kinetic energy becomes electrostatic potential energy:

K = \frac{1}{4 π ε _{0}} \frac{Z e ^{2}}{r _{0}} \Rightarrow r_{0} = \frac{k Z e ^{2}}{K} .

Here

K = 1.6 MeV = 2.56 \times 1 0^{- 13} J

.

r_{0} = \frac{( 9 \times 1 0 ^{9} ) ( 79 ) ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{2.56 \times 1 0 ^{- 13}} = \frac{1.82 \times 1 0 ^{- 26}}{2.56 \times 1 0 ^{- 13}} \approx 7.1 \times 1 0^{- 14} m .

r_{0} \approx 7.1 \times 1 0^{- 14} m

CBSE Class 12 Board · 2024 · Subjective

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