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There is a stream of neutrons with a kinetic energy of 00.03270327 eV. If the half life of neutrons is 700700 seconds, what fraction of neutrons will decay before they travel a distance of 1010 m ?

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Radhika Batra , 10 Years ago
Grade 11
anser 1 Answer
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Kevin Nash

Last Activity: 10 Years ago

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Hello Student,
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Please find the answer to your question
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K.E. = 0.0327 eV = 0.327 x 1.6 x 10-19 J
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½ mn v2n = 0.0327 x 1.6 x 10-19
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⇒ vn [2 x 0.0327 x 1.6 x 10-19/ 1.675 x 10-27]1/2
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⇒ vn = 0.25 x 104 m/s
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Time taken by the neutron to travel 10 m will be
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t = d / vn = 10 / 0.25 x 104 = 4 x 10-3 s
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Let the number of neutron initially be a.
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λ = 0.693 / t1 / 2 = 0.693 / 700 s-1
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We know that
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t = 2.303 / λ log a / a – x
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⇒ 4 x 10-3 / 2.303 x 0.693 / 700 = 1=log10 a / a – x
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⇒ log10 a / a – x = 1.72 x 10-6 ⇒ a / a – x = 1.000004
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⇒ x / a = 3. 96 x 10-6
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Thanks
Kevin Nash
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