A neutron moving with a speed 'v' mak | vedclass

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Question

For inelastic collision $\mathrm{v}^{\prime}=\frac{\mathrm{m}_{1}}{\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)} \mathrm{v}$

$=\frac{1}{(1+1)} v=\fra

Vedclass · Accepted Answer

$20.4$

Vedclass · Answer

For inelastic collision $\mathrm{v}^{\prime}=\frac{\mathrm{m}_{1}}{\left(\mathrm{m}_{1}+\mathrm{m}_{2}ight)} \mathrm{v}$

$=\frac{1}{(1+1)} v=\frac{v}{2}$

$\mathrm{n} ightarrow \mathrm{v}(\mathrm{H}) \quad$            Before

$(n)(H) ightarrow \frac{v}{2} \quad$         After Loss in $K.E.$

$=\frac{1}{2} m v^{2}-\frac{1}{2}(2 m)\left(\frac{v}{2}ight)^{2}=\frac{1}{4} m v^{2}$

$K.E.$ lost is used to jump from $1\,st$ orbit to $2\,nd$ orbit $\Delta \mathrm{K} . \mathrm{E}=10.2 \mathrm{eV}$

$\Rightarrow \frac{1}{4} m v^{2}=10.2$

Minimum $K.E.$ of neutron for inelastic collision

$\frac{1}{2} m v^{2}=2 	imes 10.2=20.4 \mathrm{eV}$

A neutron moving with a speed $'v'$ makes a head on collision with a stationary hydrogen atom in ground state . The minimum kinetic energy of the neutron for which inelastic collision will take place is....$eV$

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