Question

A radioactive element $ _{90}X^{238} $ decays into $ _{83}Y^{222} $ . then the number of $ \beta $ -panicles emitted are :

• A. 1
• B. 2
• C. 4
• D. 6

Answer 1

Answer: (A)

Emission of one $\beta$-particle increases the atomic number by 1 unit.
The given reaction can be shown as
${ }_{90} X^{238} \rightarrow{ }_{83} Y^{222}$
Change in mass number $=238-222=16$
If original nucleus decays an $\alpha$-particle it will reduce the initial mass by $4$ unit and atomic number by $2$ unit.
So, $\alpha$-pariticles emitted
$=\frac{16}{4}=4$
Change in atomic number $=90-2 \times 4=82$
But on emission of one $\beta$-particle, there is always an increase in atomic number by $1$ .
So, no. of $\beta$-particles $=83-82=1$