An alpha particle of energy 5MeV is scattered through 180∘ by a found uramiam nucleus . The distance of appreach is of the order of

Answer: CSolution: (c ) One point is ((92)235U) nranium nucleus ∴Q1=92e The other point charge is a particle ∴Q2=+2e Here the loss in K.E.=Ga∈∈P.E.(tilla−partic≤sthedistanced) ⇒21mν2=krq1q2⇒r=k21mν22q1q2 ∴r=5×1.6×10−139×109×2×1.6×10−19×92×1.6×10−19 =529.92×10−16m =592.92×10−14cm=5.2992×10−12cm

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An alpha particle of energy {5}{M}{e}{V} is scattered through

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An alpha particle of energy $5 M e V$ is scattered through $180^{\circ}$ by a found uramiam nucleus . The distance of appreach is of the order of

1 \overset{˚}{A}

10^{- 10} c m

10^{- 12} c m

10^{- 15} c m

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An alpha particle of energy

5 M e V

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180^{\circ}

by a found uramiam nucleus . The distance of appreach is of the order of

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Text solutionVerified

Answer: CSolution: (c ) One point is

(_{(92)}^{235} U)

nranium nucleus

∴ Q_{1} = 92 e

The other point charge is a particle

∴ Q_{2} = + 2 e

Here the loss in

K . E . = G a \in\in P . E . (t i l l a - p a r t i c \leq s t h e d i s tan c e d)

\Rightarrow \frac{1}{2} m ν^{2} = k \frac{q _{1} q _{2}}{r} \Rightarrow r = k \frac{2 q _{1} q _{2}}{\frac{1}{2} m ν ^{2}}

∴ r = \frac{9 \times 10 ^{9} \times 2 \times 1.6 \times 10 ^{- 19} \times 92 \times 1.6 \times 10 ^{- 19}}{5 \times 1.6 \times 10 ^{- 13}}

= 529.92 \times 10^{- 16} m

= 592.92 \times 10^{- 14} c m = 5.2992 \times 10^{- 12} c m

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Question Text	An alpha particle of energy $5 M e V$ is scattered through $180^{\circ}$ by a found uramiam nucleus . The distance of appreach is of the order of
Answer Type	Text solution:1
Upvotes	150

An alpha particle of energy 5MeV is scattered through 180∘ by a found uramiam nucleus . The distance of appreach is of the order of

Text solutionVerified

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An alpha particle of energy $5 M e V$ is scattered through $180^{\circ}$ by a found uramiam nucleus . The distance of appreach is of the order of