In the uranium radioactive series, the initia | vedclass

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

Question

(D) Let the number of $\alpha$-particles emitted be $n_{\alpha}$ and the number of $\beta$-particles emitted be $n_{\beta}$.For the decay ${}^{238}_{9

Vedclass · Accepted Answer

$8$ and $6$

Vedclass · Answer

(D) Let the number of $\alpha$-particles emitted be $n_{\alpha}$ and the number of $\beta$-particles emitted be $n_{\beta}$.For the decay ${}^{238}_{92}U \longrightarrow {}^{206}_{82}Pb + n_{\alpha}({}^{4}_{2}He) + n_{\beta}({}^{0}_{-1}e)$:Equating the mass number: $238 = 206 + 4n_{\alpha} \implies 4n_{\alpha} = 32 \implies n_{\alpha} = 8$.Equating the atomic number: $92 = 82 + 2n_{\alpha} - n_{\beta}$.Substituting $n_{\alpha} = 8$: $92 = 82 + 2(8) - n_{\beta} \implies 92 = 82 + 16 - n_{\beta} \implies 92 = 98 - n_{\beta} \implies n_{\beta} = 6$.Thus, $8$ $\alpha$-particles and $6$ $\beta$-particles are emitted.

In the uranium radioactive series, the initial nucleus ${}^{238}_{92}U$ decays to the final nucleus ${}^{206}_{82}Pb$. In this process, the number of $\alpha$-particles and $\beta$-particles emitted are:

Similar Questions

An $U^{238}$ nucleus is initially at rest. It decays by emitting an $\alpha$-particle with speed $u$. Find the recoil speed of the daughter nucleus.

The radioactive decay of uranium into thorium is expressed by the equation $_{92}^{238}U \to _{90}^{234}Th + X,$ where $X$ is

Give a brief explanation about radioactivity.

$A$ radioactive nucleus ${}_{Z}^{A}X$ undergoes spontaneous decay in the sequence ${}_{Z}^{A}X \rightarrow {}_{Z-1}B \rightarrow {}_{Z-3}C \rightarrow {}_{Z-2}D$,where $Z$ is the atomic number of element $X$. The possible decay particles in the sequence are:

The number of $\alpha$ and $\beta^-$ particles emitted during the radioactive decay chain starting from ${}_{88}^{226}Ra$ and ending at ${}_{82}^{206}Pb$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module