The nucleus _{90}Th^{228} transforms into _{8 | vedclass

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

Question

(A) Let the number of $\alpha$-particles emitted be $n_{\alpha}$ and the number of $\beta$-particles emitted be $n_{\beta}$.The change in mass number

Vedclass · Accepted Answer

$4\,\alpha, 1\,\beta$

Vedclass · Answer

(A) Let the number of $\alpha$-particles emitted be $n_{\alpha}$ and the number of $\beta$-particles emitted be $n_{\beta}$.The change in mass number is given by $A - A' = 4n_{\alpha}$.$228 - 212 = 4n_{\alpha} \implies 16 = 4n_{\alpha} \implies n_{\alpha} = 4$.The change in atomic number is given by $Z - Z' = 2n_{\alpha} - n_{\beta}$.$90 - 83 = 2(4) - n_{\beta}$.$7 = 8 - n_{\beta} \implies n_{\beta} = 1$.Thus,$4$ $\alpha$-particles and $1$ $\beta$-particle are emitted.

The nucleus $_{90}Th^{228}$ transforms into $_{83}Bi^{212}$. How many $\alpha$-particles and $\beta$-particles are emitted during this process?

Similar Questions

In a radioactive decay process, the negatively charged emitted $\beta -$ particles are

Which of the following will have the highest penetrating power?

$\beta$-decay means emission of electron from

In the nuclear reaction $6C^{11} \rightarrow 5B^{11} + \beta^+ + X$,the particle $X$ is .......

In the nuclear reaction $_{92}U^{238} \to _{Z}Th^{A} + _{2}He^{4}$,the values of $A$ and $Z$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module