In the uranium radioactive series,the initial | vedclass

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

Question

(A) Let $n_{\alpha}$ be the number of $\alpha$-particles and $n_{\beta}$ be the number of $\beta$-particles emitted.For the mass number change: $238 =

Vedclass · Accepted Answer

$8 \alpha, 6 \beta$

Vedclass · Answer

(A) Let $n_{\alpha}$ be the number of $\alpha$-particles and $n_{\beta}$ be the number of $\beta$-particles emitted.For the mass number change: $238 = 206 + 4n_{\alpha} + 0n_{\beta}$.$4n_{\alpha} = 238 - 206 = 32 \implies n_{\alpha} = 8$.For the atomic number change: $92 = 82 + 2n_{\alpha} - 1n_{\beta}$.Substituting $n_{\alpha} = 8$: $92 = 82 + 2(8) - n_{\beta}$.$92 = 82 + 16 - n_{\beta} \implies 92 = 98 - n_{\beta}$.$n_{\beta} = 98 - 92 = 6$.Thus,$8$ $\alpha$-particles and $6$ $\beta$-particles are emitted.

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

Similar Questions

In $\beta^{-}$ decay, a neutron transforms into a proton within the nucleus according to the equation:
$\text{neutron} \rightarrow \text{proton} + \beta^{-} + x$
In this equation, the particle represented by '$x$' is:

Identify the particle $x$ in the following reaction: ${}_3^7Li(p, x){}_4Be^8$,where $p$ is a proton.

Which waves are used in the treatment of cancer?

When an element ${}_{90}^{232}Th$ decays into ${}_{82}^{208}Pb$, the number of $\alpha$ and $\beta^{-}$ particles emitted respectively are

$A$ nucleus of lead $Pb_{82}^{214}$ emits two electrons followed by an $\alpha$-particle. The resulting nucleus will have

Vedclass Test Series

Exam Paper Generator

Online Exam Module