A radioactive element $_{90}{X^{238}}$ decay into $_{83}{Y^{222}}$. The number of $\beta - $ particles emitted are
A radioactive element $_{90}{X^{238}}$ decay into $_{83}{Y^{222}}$. The number of $\beta - $ particles emitted are
- A
$4$
- B
$6$
- C
$2$
- D
$1$
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Alpha particles are
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- [AIPMT 1999]
In the equation ${ }_{13}^{27} Al +{ }_2^4 He \longrightarrow{ }_{15}^{30} P + X ,$ The correct symbol for $X$ is
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${U^{238}}$ decays into $T{h^{234}}$ by the emission of an $\alpha - $ particle. There follows a chain of further radioactive decays, either by $\alpha - $ decay or by $\beta $ - decay. Eventually a stable nuclide is reached and after that, no further radioactive decay is possible. Which of the following stable nuclides is the and product of the ${U^{238}}$ radioactive decay chain
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Three $\alpha - $ particles and one $\beta - $ particle decaying takes place in series from an isotope $_{88}R{a^{238}}$. Finally the isotope obtained will be
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A nucleus $_Z{X^A}$ emits $3 \alpha$ - particles and $5 \beta$ particle. The ratio of total neutrons and protons in the final nucleus is
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