Q12 / 150 · JEE Main 2006 · Chemistry

In the transformation of ${ }_{92}^{238} \mathrm{U}$ to ${ }_{92}^{234} \mathrm{U}$, if one emission is an $\alpha$-particle, what should be the other emission(s)?

  • Correct answer
    A
    Two $\beta^{-}$
  • B
    Two $\beta^{-}$and one $\beta^{+}$
  • C
    One $\beta^{-}$and one $\gamma$
  • D
    One $\beta^{+}$and one $\beta^{-}$

Solution

${ }_{92}^{238} \mathrm{U} \longrightarrow{ }_{92}^{234} \mathrm{U}+{ }_2^4 \mathrm{He}+2{ }_{-1}^0 \mathrm{e}$

Hence, the correct answer is option: 'A'.

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