SATHEE: Atoms And Nuclei Question 277

Atoms And Nuclei Question 277

Question: An $α$ -particle of energy 5 MeV is scattered through $180^{\circ}$ by a fixed uranium nucleus. The distance of closest approach is of the order of

Options:

A) $10^{- 12} c m$

B) $10^{- 10} c m$

C) $10^{- 14} c m$

D) $10^{- 15} c m,$

Show Answer

Answer:

Correct Answer: A

Solution:

Distance of closest approach $Energy, E = 5 \times 10^{6} \times 1.6 \times 10^{- 19} J .$ $r_{0} = \frac{Z e (2 e)}{4 π ε_{0} (\frac{1}{2} m v^{2})}$
$∴ r_{0} = \frac{9 \times 10^{9} \times (92 \times 1.6 \times 10^{- 19}) \times (2 \times 1.6 \times 10^{- 19})}{5 \times 10^{6} \times 1.6 \times 10^{- 19}}$
$\Rightarrow r = 5.2 \times 10^{- 14} m = 5.3 \times 10^{- 12} c m .$

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Atoms And Nuclei Question 277

Question: An α -particle of energy 5 MeV is scattered through 180∘ by a fixed uranium nucleus. The distance of closest approach is of the order of

Options:

Answer:

Solution:

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Question: An $α$ -particle of energy 5 MeV is scattered through $180^{\circ}$ by a fixed uranium nucleus. The distance of closest approach is of the order of