1.

Over what distance in free space will the intensity of a 5eV neutron beam be reduced by a factor one half ?

Answer»

23808 km
15070 km
10208 km
5028 km

Solution :`1/2 MV^(2)=5eV or`
`v=sqrt((2 xx 5 xx 1.6 xx 10^(-19))/(1.67 xx 10^(-27)) =31.0"km//sec"`
During a time to half PERIOD T=12.8 min, half the neutron would have decayed from the beam. The distance travelled by undecayed NEUTRONS during this time is `S=v xx T=31.0 xx 12.8 xx 60km`
=23808 km


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