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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 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

When the atomic number Z of the nucleus increases

Knowledge Check

  • A metal M of equivalent mass E froms an oxide of molecular formula M_(x)O_(y) . The atomic mass of the metal is given by the correct equation

    A
    `2E(y//x)`
    B
    xyE
    C
    E/y
    D
    y/E

  • The binding energy per nucleon is almost constant for the nuclei having atomic mass number……….

    A
    `30 lt A lt 170`
    B
    `30 lt A lt 240`
    C
    `170 lt A lt 230`
    D
    `156 lt A lt 192`

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