What is the respective number of alpha and be | vedclass

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

Question

(D) In an $\alpha$-decay,the mass number decreases by $4$ and the atomic number decreases by $2$. In a $\beta$-decay,the mass number remains unchanged

Vedclass · Accepted Answer

$8$ and $6$

Vedclass · Answer

(D) In an $\alpha$-decay,the mass number decreases by $4$ and the atomic number decreases by $2$. In a $\beta$-decay,the mass number remains unchanged and the atomic number increases by $1$.Let $n_{\alpha}$ be the number of $\alpha$-particles and $n_{\beta}$ be the number of $\beta$-particles emitted.The change in mass number is given by: $200 - 168 = 4 	imes n_{\alpha}$.Therefore,$n_{\alpha} = \frac{32}{4} = 8$.The change in atomic number is given by: $90 - 80 = 2 	imes n_{\alpha} - 1 	imes n_{\beta}$.Substituting $n_{\alpha} = 8$: $10 = 2(8) - n_{\beta}$.$10 = 16 - n_{\beta}$,which gives $n_{\beta} = 6$.Thus,the number of $\alpha$ and $\beta$ particles emitted are $8$ and $6$ respectively.

What is the respective number of $\alpha$ and $\beta$ particles emitted in the following radioactive decay: $_{90}X^{200} \to _{80}Y^{168}$?

Similar Questions

During a $\beta^{-}$-decay,

Three $\alpha$-particles and one $\beta$-particle decay take place in series from an isotope $_{88}Ra^{238}$. Finally,the isotope obtained will be:

For the given reaction, the particle $X$ is $_6C^{11} \to _5B^{11} + \beta^+ + X$.

In a radioactive decay chain reaction,${ }_{90}^{230} Th$ nucleus decays into ${ }_{84}^{214} Po$ nucleus. The ratio of the number of $\alpha$ to number of $\beta^{-}$ particles emitted in this process is. . . . .

$A$ 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?

Vedclass Test Series

Exam Paper Generator

Online Exam Module