Atoms & Nuclei

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21 of 239

Question 21 of 239

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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Answer: (A) 20.4 eV

Let velocity before collision v from momentum conservation mv=(m+m)v1mv=(m+m){v}^{1}

v1=v2{v}^{1}=\dfrac{v}{2}

Loss in K.E. =12mv2−12(2m)(v2)2=\dfrac{1}{2}m{v}^{2}-\dfrac{1}{2}\left(2m\right){\left(\dfrac{v}{2}\right)}^{2}

=14mv2=\dfrac{1}{4}m{v}^{2}

K.E. lost is used to jump from 1st{1}^{st} orbit to 2nd{2}^{nd} orbit ∆K.E.=10.2eV∆K.E.=10.2eV

⇒14mv2=10.2\Rightarrow \dfrac{1}{4}m{v}^{2}=10.2

12mv2=2×10.2=20.4eV\dfrac{1}{2}m{v}^{2}=2\times 10.2=20.4eV

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