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  4. A radioactive element 90X238 decays into 83 Upsilon 222. The number of β-particle emitted are:

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Q. A radioactive element $ _{90}{{X}^{238}} $ decays into $ _{83}{{\Upsilon }^{222}}. $ The number of $\beta$-particle emitted are:

JIPMERJIPMER 2001Nuclei

Solution:

$ \alpha $ -particle are emitted $ =\frac{238-222}{4}=4 $ The atomic number is decreased $ 90-4\times 2=82 $ As atomic number of $ {}_{83}{{Y}^{222}}, $ so atomic number is increased by 1, therefore one $ \beta $ -particle is emitted.