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Q.

An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of the closest approach is of the order of

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a

10-15 cm

b

10-13 cm

c

10-12 cm

d

10-19 cm

answer is C.

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Detailed Solution

Use E=14πε0(Ze)(e)r0 5×1.6×10-13=9×109×2×92×1.6×10-192r0 r0=5.3×10-14m≈10-12cm
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