A neutron of energy 2 MeV 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.

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

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

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$m v l_{\min} = L \Rightarrow \sqrt{2 m_{n e u t r o n} K_{n e u t r o n}} = 4 \times 10^{- 34}$

$\Rightarrow l_{\min} = 1.25 \times 10^{- 14} m = 125 \times 10^{- 16} m$

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