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5.work,energy,power and collision — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysics5.work,energy,power and collision1 Mark
MCQ
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....$eV$
  • ✓
    $20.4$
  • B
    $10.2$
  • C
    $12.1$
  • D
    $16.8$
✓

Answer

Correct option: A.
$20.4$
a
For inelastic collision $\mathrm{v}^{\prime}=\frac{\mathrm{m}_{1}}{\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)} \mathrm{v}$

$=\frac{1}{(1+1)} v=\frac{v}{2}$

$\mathrm{n} \rightarrow \mathrm{v}(\mathrm{H}) \quad$            Before

$(n)(H) \rightarrow \frac{v}{2} \quad$         After Loss in $K.E.$

$=\frac{1}{2} m v^{2}-\frac{1}{2}(2 m)\left(\frac{v}{2}\right)^{2}=\frac{1}{4} m v^{2}$

$K.E.$ 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

$\frac{1}{2} m v^{2}=2 \times 10.2=20.4 \mathrm{eV}$

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