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After emission of one alpha particle fol...

After emission of one `alpha` particle followed by one `beta`-particle from `""_(92)^(238)X`, the number of neutrons in the atom will be

A

142

B

146

C

144

D

143

Text Solution

Verified by Experts

The correct Answer is:
D
`._(92)X^(238) overset(-alpha)rarr ._(90)Y^(234)overset(-beta)rarr ._(91)Z^(234)`
no. of neutrons `= 234 - 91 = 143`
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Knowledge Check

  • After the emission of one alpha - particle followed by two beta -particles from "" _(92) ^(238) U, the number of neutrons in the newly formed nucleus is

    A
    140
    B
    142
    C
    144
    D
    146

  • The emission of beta particle is from

    A
    The valence shell of an atom
    B
    The inner shell of an atom
    C
    The nucleus due to the nuclear conversion `:`
    Proton `rarr` neutron`+` electron
    D
    The nucleus due to the nuclear conversion `:`
    neutron `rarr` proton `+ ` electron

  • After the emission of alpha -particle from the atom X_92^238 , the number of neutrons in the atom will be

    A
    138
    B
    140
    C
    144
    D
    150

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