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  • A neutron moving with  peed <img alt="\mathrm{v}" src="https://entrancecorner.oncodecogs.com/gif.

A neutron moving with \mathrm{5} peed \mathrm{v} makes a head-on collision with a hydrogen atom in ground state kept at rest. The minimum kinetic energy of the neutron for which inelastic collision may take place - The mass of the neutron \mathrm{=} mass of the hydrogen \mathrm{=1.67 \times 10^{-27}kg}

Option: 1

\mathrm{\frac{1}{2} m v^2>\Delta E}


Option: 2

\mathrm{\frac{1}{2} m v^2<2 \Delta E}


Option: 3

\mathrm{\frac{1}{2} m v^2>2 \Delta E}


Option: 4

\mathrm{\frac{1}{2} \min ^2<\Delta E}


Answers (1)

best_answer

Using conservation of liner momentum and energy - 
\mathrm{mv=mv_{1}+mv_{2}\ \ \ ........\left ( 1 \right )}
\mathrm{\text { and } \frac{1}{2} m c^2=\frac{1}{2} m \theta_1^2+\frac{1}{2} m c_2^2+\Delta E\ \ \ \\ \ ......... \text { (2) }}
\mathrm{\text { from (1) } \quad v^2=v_1^2+v_2^2+2 v_1 v_2}
\mathrm{\text { from (2) } \quad v^2=v_1^2+v_2^2+\frac{2\Delta E}{m}}
Thus, \mathrm{2 v_1 v_2=\frac{2 \Delta E}{m}}
Hence, \mathrm{\left(v_1-k_2\right)^2=\left(v_1+v_2\right)^2-4 v_1 v_2}
\mathrm{=v^2-\frac{4 \Delta E}{m}}
As \mathrm{(v_{1}-v_{2})} must be real - 
\mathrm{v^2-\frac{4 \Delta E}{m} \geqslant 0}
\mathrm{\frac{1}{2} m v^2>2 \Delta E}

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Gunjita

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