JEE Main · 2018 · 08 Apr 2018 · Offline · Conservation of Momentum

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It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is

p_{d}

; while for its similar collision with carbon nucleus at rest, fractional loss of energy is

p_{c}

. The values of

p_{d}

and

p_{c}

are respectively:

(A) $(0, 1)$
(B) $(.89, .28)$ ✓ Correct
(C) $(.28, .89)$
(D) $(0, 0)$

Answer: (B)

Solution

For a head-on elastic collision of a neutron of mass

1

with a stationary nucleus of mass

A

, the fraction of energy the neutron keeps is

(\frac{A - 1}{A + 1})^{2}

, so the fractional loss is

1 - (\frac{A - 1}{A + 1})^{2}

. For deuterium,

A = 2

, the loss is

1 - (\frac{1}{3})^{2} = 0.89

, and for carbon,

A = 12

, it is

1 - (\frac{11}{13})^{2} \approx 0.28

, so

p_{d} = 0.89

and

p_{c} = 0.28

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option

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