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The mass of nucleus .(z)X^(A) is less th...

The mass of nucleus `._(z)X^(A)` is less than the sum of the masses of `(A-Z)` number of neutrons and `Z` number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass `M` can break into two light nuclei of mass `m_(1)` and `m_(2)` only if `(m_(1)+m_(2)) lt M`. Also two light nuclei of massws `m_(3)` and `m_(4)` can undergo complete fusion and form a heavy nucleus of mass `M` ''only if `(m_(3)+m_(4)) gt M`''. The masses of some neutral atoms are given in the table below.
`|{:(._(1)^(1)H,1.007825u,._(1)^(2)H,2.014102u,),(._(1)^(3)H,3.016050u,._(2)^(4)H,4.002603u,),(._(3)^(6)Li,6.015123u,._(3)^(7)Li,7.016004u,),(._(30)^(70)Zn,69.925325u,._(34)^(82)Se,81.916709u,),(._(64)^(152)Gd,151.91980u,._(82)^(206)Pb,205.97445u,),(._(83)^(209)Bi,208.980388u,._(84)^(210)Po,209.982876u,):}|`
The correct statement is

A

The nucleus `._(3)^(6)Li` can emit an alpha particle

B

The nucleus `._(84)^(210)Po` can emit a proton

C

Deutron and alpha particle can undergo complete fusion.

D

The nuclei `._(30)^(70)Zn` and `._(34)^(82)Se` can usdergo complete fusion.

Text Solution

Verified by Experts

The correct Answer is:
C

`(Q)/(C^(2))= Delta m=M_("reactants")-M_("Products")`
if `(Q)/(C^(2)) gt0 , alpha`-decay is possible
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