An α -particle of energy 5 MeV is scattered through 180^ by a fixed uranium nucleus ( Z = 92 ). The distance of the closest approach is of the order o - Physics MCQ AIEEE 2004 2026 | ExamTest.live

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An α -particle of energy 5 MeV is scattered through 18 0^{\circ} by a fixed uranium nucleus (Z = 92). The distance of the closest approach is of the order of:

Solution & Step-by-step Explanation

At the distance of closest approach (r_{0}), the initial kinetic energy is converted to potential energy. K = \frac{1}{4 π ϵ _{0}} \frac{q _{1} q _{2}}{r _{0}} Given: K = 5 MeV = 5 \times 1 0^{6} \times 1.6 \times 1 0^{- 19} J q_{1} = 2 e, q_{2} = 92 e $ r_{0} = \frac{1}{4 π ϵ _{0}} \frac{( 2 e ) ( 92 e )}{K} = \frac{( 9 \times 1 0 ^{9} ) \times 184 \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{5 \times 1 0 ^{6} \times 1.6 \times 1 0 ^{- 19}} $ r_{0} \approx \frac{9 \times 1 0 ^{9} \times 184 \times 1.6 \times 1 0 ^{- 19}}{5 \times 1 0 ^{6}} \approx 5.3 \times 1 0^{- 14} m r_{0} = 5.3 \times 1 0^{- 12} cm The order of magnitude is 1 0^{- 12} cm .

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An α -particle of energy 5 MeV is scattered through 18 0^{\circ} by a fixed uranium nucleus (Z = 92). The distance of the closest approach is of the order of:

1 \overset{˚}{A}

1 0^{- 10} cm

1 0^{- 12} cm

1 0^{- 15} cm

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