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 perfactly inelastic collision will take place is :

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Correct answer: A. 20.4 eV

Let velocity before collision v from momentum conservation

mv = ( m + m ){v}^{1}

{v}^{1} = \dfrac{v}{2}

Loss in K.E.

= \dfrac{1}{2}m{v}^{2} - \dfrac{1}{2}\left(2m\right){\left(\dfrac{v}{2}\right)}^{2}

= \dfrac{1}{4}m{v}^{2}

K.E. lost is used to jump from

{1}^{st}

orbit to

{2}^{nd}

orbit

∆ K . E . = 10.2eV

⇒ \dfrac{1}{4}m{v}^{2} = 10.2

\dfrac{1}{2}m{v}^{2} = 2 \times 10.2 = 20.4eV

A neutron moving with a speed 'v' makes a head on collision with a...

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