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` alpha`-particles and ` beta`-particles

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Calculate the atomic number and mass number of the resulting element, when ""_(92)^(235)U emits 6 alpha -particles and 4 beta -particles.

In a radioactive decay chain, the initial nucleus is overset(232)(90).^(Th) . At the end there are 6 alpha -particles and 4 beta -particles which are emitted. If the end nucleus is overset(A)(Z).^(X),A and Z are given by :

Knowledge Check

  • ._(92)U^(238) emits 8 alpha- particles and 6 beta- particles. The n//p ratio in the product nucleus is

    A
    `(62)/(41)`
    B
    `(60)/(41)`
    C
    `(61)/(42)`
    D
    `(62)/(42)`

  • A radioactive nucleus ._(Z) X^(A) emite 3 alpha- particles and 5 beta - particles . The ratio of number of neutrons to that of protons in the producet nucleus will be :

    A
    `(A - Z - 12)/( Z-6)`
    B
    ` (A-Z)/(Z-1)`
    C
    ` (A-Z-11)/( Z-6)`
    D
    ` (A-Z -11)/( Z-1)`

  • In a radioactive decay chain, the initial nucleus is overset(232)(90).^(Th) . At the end there are 6 alpha -particles and 4 beta -particles which are emitted. If the end nucleus is overset(A)(Z).^(X),A and Z are given by :

    A
    `A=208 , Z=80`
    B
    `A=202 , Z=80`
    C
    `A=208 , Z=82`
    D
    `A=200 , Z=81`

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    A nucleus ._(Z)X^(A) emits 2 alpha - particles and 3 beta - particles. The ratio of total protons and neutrons in the final nucleus is :

    ._(92)^(238)U emits 8 alpha -particles and 6-beta -particles. The neutron/proton ratio in the product nucleus is

    Successive emission of an alpha -particle and two beta -particles by an atom of an element result in the formation of its

    An element with atomic number 84 and mass number 218 loses one alpha- particle and two beta- particles in three successive stages, the resulting element will have

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