A neutron makes a head on elastic collision with a stationary deuteron. The fractional energy loss of the neutron in the collision is :-

Question

A neutron makes a head on elastic collision with a stationary deuteron. The fractional energy loss of the neutron in the collision is :- (A) 16/82 (B) 8/9 (C) 8/27 (D) 2/3

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-bizloajqiikumyag.jpg" /> u=v1+2v2...(1) e=1=(- (v1 - v2)/u) ∴ u=v2-v1...(2) solving v1=(- u/3),v2=(2 u/3) fraction energy loss of neutron ((1/2) m u2 - (1/2) m (- (u/3))2/(1)2 m u2)=(8/9)

Q. A neutron makes a head on elastic collision with a stationary deuteron. The fractional energy loss of the neutron in the collision is :-

$16/82$

$8/9$

$8/27$

$2/3$

$u = v_{1} + 2 v_{2} ... (1)$
$e = 1 = \frac{- ( v _{1} - v _{2} )}{u}$
$∴ u = v_{2} - v_{1} ... (2)$
solving $v_{1} = \frac{- u}{3}, v_{2} = \frac{2 u}{3}$
fraction energy loss of neutron
$\frac{\frac{1}{2} m u ^{2} - \frac{1}{2} m ( - \frac{u}{3} ) ^{2}}{\frac{1}{2} m u ^{2}} = \frac{8}{9}$

Q. A neutron makes a head on elastic collision with a stationary deuteron. The fractional energy loss of the neutron in the collision is :-

16/82

8/9

8/27

2/3

u=v1​+2v2​...(1) e=1=u−(v1​−v2​)​ ∴u=v2​−v1​...(2) solving v1​=3−u​,v2​=32u​ fraction energy loss of neutron 21​mu221​mu2−21​m(−3u​)2​=98​

$16/82$

$8/9$

$8/27$

$2/3$

$u = v_{1} + 2 v_{2} ... (1)$
$e = 1 = \frac{- ( v _{1} - v _{2} )}{u}$
$∴ u = v_{2} - v_{1} ... (2)$
solving $v_{1} = \frac{- u}{3}, v_{2} = \frac{2 u}{3}$
fraction energy loss of neutron
$\frac{\frac{1}{2} m u ^{2} - \frac{1}{2} m ( - \frac{u}{3} ) ^{2}}{\frac{1}{2} m u ^{2}} = \frac{8}{9}$