If _{92}U^{238} undergoes successively 8 alph | vedclass

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

Question

(B) The initial nucleus is $_{92}U^{238}$.An $\alpha$-decay reduces the mass number $A$ by $4$ and the atomic number $Z$ by $2$.$A$ $\beta$-decay does

Vedclass · Accepted Answer

$_{82}Pb^{206}$

Vedclass · Answer

(B) The initial nucleus is $_{92}U^{238}$.An $\alpha$-decay reduces the mass number $A$ by $4$ and the atomic number $Z$ by $2$.$A$ $\beta$-decay does not change the mass number $A$ and increases the atomic number $Z$ by $1$.After $8 \alpha$-decays,the new mass number $A' = 238 - (8 	imes 4) = 238 - 32 = 206$.The new atomic number $Z' = 92 - (8 	imes 2) + (6 	imes 1) = 92 - 16 + 6 = 82$.The element with atomic number $82$ is Lead $(Pb)$.Therefore,the resulting nucleus is $_{82}Pb^{206}$.

If $_{92}U^{238}$ undergoes successively $8 \alpha$-decays and $6 \beta$-decays,then the resulting nucleus is

Similar Questions

$A$ nucleus $_Z^AX$ emits an $\alpha$-particle. The resultant nucleus then emits a ${\beta ^ + }$ particle. What will be the respective atomic number and mass number of the final nucleus?

When a radioactive isotope ${}_{88}^{228}Ra$ decays in series by the emission of $3$ alpha particles and $1$ beta particle,the isotope finally formed is

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

The energy spectrum of $\beta$-particles [number $N(E)$ as a function of $\beta$-energy $E$] emitted from a radioactive source is:

Curie is a unit of

Vedclass Test Series

Exam Paper Generator

Online Exam Module