The A_{92}U^{238} nucleus decays to a Pb^{214 | 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 $n_{\beta}$.For the mass number: $

Vedclass · Accepted Answer

$6$ and $2$

Vedclass · Answer

(A) Let the number of $\alpha$-particles emitted be $n_{\alpha}$ and the number of $\beta^{-}$-particles emitted be $n_{\beta}$.For the mass number: $238 = 214 + 4n_{\alpha} \implies 4n_{\alpha} = 24 \implies n_{\alpha} = 6$.For the atomic number: $92 = 82 + 2n_{\alpha} - 1n_{\beta}$.Substituting $n_{\alpha} = 6$: $92 = 82 + 2(6) - n_{\beta} \implies 92 = 82 + 12 - n_{\beta} \implies 92 = 94 - n_{\beta} \implies n_{\beta} = 2$.Thus,the number of $\alpha$-particles is $6$ and the number of $\beta^{-}$-particles is $2$.

The $A_{92}U^{238}$ nucleus decays to a $Pb^{214}_{82}$ nucleus. The number of $\alpha$ and $\beta^{-}$ particles emitted are:

Similar Questions

In the emission of $\gamma$-rays from a nucleus:

Assertion: Cobalt-$60$ is useful in cancer therapy.
Reason: Cobalt-$60$ is a source of $\gamma$-radiations capable of killing cancerous cells.

Who discovered radioactivity?

Radioactivity was discovered by

$\beta$-decay means emission of electron from

Vedclass Test Series

Exam Paper Generator

Online Exam Module