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  4. Distance in free space at which intensity of 5 eV neutron beam reduces to half will be nearly: (Take half-life of the neutron =12.8 min )

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Q. Distance in free space at which intensity of $5 \, eV$ neutron beam reduces to half will be nearly: (Take half-life of the neutron $=12.8 \, min$ )

NTA AbhyasNTA Abhyas 2022

Solution:

The kinetic energy of the neutron beam $5 \, eV=\frac{1}{2} \, \text{mv}^{2}$
$\frac{1}{2}\left(1.67 \times \left(10\right)^{- 27}\right)v^{2}=5\times 1.6\times \left(10\right)^{- 19}$
$\Rightarrow v=31 \, $ $kms^{- 1}$
distance travelled in one half-life $d=vt=31\times 10^{3}\times 12.8\times 60m\sim eq24000 \, km \, $