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  • In free space the intensity of 5 eV neutron beam is reduced by a factor of one half. Half life is  min. The distance tr

In free space the intensity of 5 eV neutron beam is reduced by a factor of one half. Half life is t_{1 / 2}=12.8 min. The distance travelled by neutron beam is-

Option: 1

2800 km


Option: 2

23800 km 


Option: 3

28 km


Option: 4

2 km


Answers (1)

best_answer

Speed of the neutrons in beam is

\frac{1}{2} \mathrm{mv}^2=\mathrm{K}=5 \mathrm{eV}

v=\sqrt{\frac{2(5) \times 1.6 \times 10^{-19}}{1.67 \times 10^{-27}}}

\mathrm{v}=31 \mathrm{~km} / \mathrm{sec}

During a time of  t_{1 / 2}=12.8 \mathrm{~min}, half the neutrons will have decayed from the beam. The distance travelled by the undecayed during this time is 

\mathrm{d}=\mathrm{vt}=(31 \mathrm{~km} / \mathrm{s})(12.8 \mathrm{~min})(60 \mathrm{~s} / \mathrm{min})

=23800 \mathrm{~km}

Posted by

Kuldeep Maurya

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