The number of alpha- and beta- particles emit | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The nuclear transformation is $_{90}^{232}Th \xrightarrow{} _{82}^{208}Pb + x(\alpha) + y(\beta)$.Step $1$: Calculate the number of $\alpha-$ part

Vedclass · Accepted Answer

$6, 4$

Vedclass · Answer

(D) The nuclear transformation is $_{90}^{232}Th \xrightarrow{} _{82}^{208}Pb + x(\alpha) + y(\beta)$.Step $1$: Calculate the number of $\alpha-$ particles $(x)$: The change in mass number is $232 - 208 = 24$. Since each $\alpha-$ particle has a mass of $4$,$x = \frac{24}{4} = 6$.Step $2$: Calculate the number of $\beta-$ particles $(y)$: The change in atomic number is $90 - 82 = 8$. Each $\alpha-$ particle decreases the atomic number by $2$,so $6 \alpha-$ particles decrease it by $12$. Each $\beta-$ particle increases the atomic number by $1$. Thus,$90 - (6 	imes 2) + y = 82$,which gives $90 - 12 + y = 82$,so $78 + y = 82$,resulting in $y = 4$.Therefore,the number of $\alpha-$ and $\beta-$ particles emitted are $6$ and $4$ respectively.

The number of $\alpha-$ and $\beta-$ particles emitted respectively during the transformation of $_{90}^{232}Th$ to $_{82}^{208}Pb$ is:

Similar Questions

The $_{88}Ra^{226}$ is

Of the following radiations,the one most easily stopped by air is

$\beta$-particles are emitted from the atom

In the following nuclear reaction,identify $x$: $_{29}Cu^{64} \to _{28}Ni^{64} + x$

The disintegration of an isotope of sodium,$_{11}Na^{24} \to _{12}Mg^{24} + _{-1}e^0$,shown is due to

Vedclass Test Series

Exam Paper Generator

Online Exam Module