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The mass of a nucleus .(Z)^(A)X is less ...

The mass of a nucleus `._(Z)^(A)X` is less that 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 masses `m_(1)` and `m_(2)` only if `(m_(1)+m_(2)) lt M`. Also two light nuclei of masses `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)He,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,205.974455u,._(83)^(209)Bi,208.980388u,._(84)^(210)Po,209.982876u):}|`
Taking kinetic energy ( in `KeV`) of the alpha particle, when the nucleus `._(84)^(210)P_(0)` at rest undergoes alpha decay, is:

A

`5319`

B

`5422`

C

`5707`

D

`5818`

Text Solution

Verified by Experts

The correct Answer is:
A
`._(84)^(210)Po rarr ._(82)^(206)Pb+_(2)^(4)He+Q`
Total energy released `=(M_(Po)-M_(Pb)-M_(He))C^(2)`
`=[(209.982876)-(205.974455+4.002603)]xx932 MeV`
`=[0.005818]xx932 MeV=5.422376 MeV`
Kinetic energy of `alpha` particle
`=((A-4)/A)Q=(206/210)5.422376 MeV`
`=5.319 MeV=5319 KeV`
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The mass of a nucleus ._(Z)^(A)X is less that 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 masses m_(1) and m_(2) only if (m_(1)+m_(2)) lt M . Also two light nuclei of masses 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)He,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,205.974455u,._(83)^(209)Bi,208.980388u,._(84)^(210)Po,209.982876u):}| The correct statement is:

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