A neutron of energy 2MeV and mass 1.6× 10- 27kg passes a proton at such a distance that the angular momentum of neutron relative to proton approximately equals 4× 10- 34Js . The distance of closest approach neglecting the interaction between particles is given by α × 10- 16m. Find the value of α .

Question

A neutron of energy 2MeV and mass 1.6× 10- 27kg passes a proton at such a distance that the angular momentum of neutron relative to proton approximately equals 4× 10- 34Js . The distance of closest approach neglecting the interaction between particles is given by α × 10- 16m. Find the value of α . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

mvlmin=L ⇒ √2 mneutron Kneutron× lm i n=4× 10- 34 ⇒ lmin=1.25× 10- 14m=125× 10- 16m

mvlmin​=L ⇒2mneutron​Kneutron​​×lmin​=4×10−34 ⇒lmin​=1.25×10−14m=125×10−16m

$m v l_{min} = L$
$\Rightarrow 2 m_{n e u t ro n} K_{n e u t ro n} \times l_{min} = 4 \times 1 0^{- 34}$
$\Rightarrow l_{min} = 1.25 \times 1 0^{- 14} m = 125 \times 1 0^{- 16} m$