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  4. 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 .

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Q. A neutron of energy $2MeV$ 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 $\alpha \times 10^{- 16}m.$ Find the value of $\alpha $ .

NTA AbhyasNTA Abhyas 2022

Solution:

$mvl_{min}=L$
$\Rightarrow \sqrt{2 m_{neutron} K_{neutron}}\times l_{m i n}=4\times 10^{- 34}$
$\Rightarrow l_{min}=1.25\times 10^{- 14}m=125\times 10^{- 16}m$