1.

Consider the following nuclear fission reaction `._(88)Ra^(226) to ._(86)Rn^(222)+._(2)He^(4)+Q`In the fission reaction. Kinetic energy of `alpha`-particle is 4.44 MeV. Find the energy emitted as `gamma`-radiation in keV in this reaction. `m(._(88)Ra^(226))=226.005` amu `m(._(86)Rn^(222))=222.000` amu

Answer» (a) Alpha particle decay of `""_(88)^(226)Ra` emits a helium nucles. As a result, its mass number reduces to (226-4) 222 and its atomic number reduces to (88-2)86. This is shown in the following nuclear reaction.
`""_(88)^(226)Rato""_(86)^(222)Ra+""_(2)^(4)He`
Q-value of
emitted `alpha`-particle = (Sum of initial mass - Sum of final mass)
`c^(2)`
Where, c= speed of light
It is given that:
`m(""_(88)^(226)Ra)=226.02540u`
`m(""_(86)^(222)Rn)=222.01750u`
`m(""_(2)^(4)He)=4.002603u`
Q-value = [226.02540-(222.01750+4.002603)] u `c^(2)`
= 0.005297 u `c^(2)`
But 1 u = 931.5 `MeV//C^(2)`
`thereforeQ=0.005297xx931.5~~4.94 MeV`
Kinetic energy of the `alpha`-particle
`=(("Mass number after decay")/("Mass number before decay"))xxQ`
`=(222)/(226)xx4.94=4.85MeV`
(b) Alpha particle decay of `(""_(86)^(220)Rn)` is shown by the following nuclear reaction.
`""_(86)^(220)Rnto""_(84)^(216)"Po"+""_(2)^(4)He`
It is given that:
Mass of `(""_(86)^(220)Rn)` = 220.01137 u
Mass of `(""_(84)^(216)Po)` = 216.00189 u
`therefore`Q-value= [220.01137-(216.00189+4.00260)]`xx` 931.5
`~~641` MeV
Kinetic energy of the `alpha`-particle `=((220-4)/(220))xx6.41`
=6.29 MeV


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