When $_{88}{\rm{R}}{{\rm{a}}^{236}}$ decays in a series by emission of $3\,{\rm{\alpha }}$-particles and one ${\rm{\beta }}$-particle, isotope $X$ formed is

JEE PHYSICS MCQ: NUCLEI

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Application +4 | -1

When

_{88}\mathrm{R}{\mathrm{a}^{236}}

decays in a series by emission of

3\,\mathrm{\alpha }

-particles and one

\mathrm{\beta }

-particle, isotope

X

formed is

View answer and explanation

Correct option: A

_{83}{X^{224}}

Option A (

_{83}{X^{224}}

) is correct because after emitting 3 alpha particles and one beta particle, the resulting isotope has an atomic number of 83 and a mass number of 224.

JEE PHYSICS MCQ: NUCLEI

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