The end product of the decay of { }_{90} Th^ | 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 $ x $ and the number of $ \beta $-particles emitted be $ y $. The decay reaction is: $ { }_{90}

Vedclass · Accepted Answer

$ 6, 4 $

Vedclass · Answer

(D) Let the number of $ \alpha $-particles emitted be $ x $ and the number of $ \beta $-particles emitted be $ y $. The decay reaction is: $ { }_{90} Th^{232} \rightarrow { }_{82} Pb^{208} + x({ }_{2} He^{4}) + y({ }_{-1} e^{0}) $. Equating the mass numbers: $ 232 = 208 + 4x \implies 4x = 24 \implies x = 6 $. Equating the atomic numbers: $ 90 = 82 + 2x - y $. Substituting $ x = 6 $: $ 90 = 82 + 2(6) - y \implies 90 = 82 + 12 - y \implies 90 = 94 - y \implies y = 4 $. Thus,$ 6 $ $ \alpha $-particles and $ 4 $ $ \beta $-particles are emitted.

The end product of the decay of $ { }_{90} Th^{232} $ is $ { }_{82} Pb^{208} $. The number of $ \alpha $ and $ \beta $ particles emitted are respectively:

Similar Questions

Which waves are used in the treatment of cancer?

The intensity of gamma radiation emitted from a source is $I_0$. After passing through a lead sheet of thickness $36 \, mm$,its intensity becomes $I_0/8$. Find the thickness of the lead sheet in $mm$ when the intensity becomes $I_0/2$.

In the radioactive disintegration series,for the process $_{92}^{238}U \xrightarrow{\alpha} X \xrightarrow{\beta^-} {}_{Z}^{A}Y$,the values of $Z$ and $A$ respectively are:

The '$rad$' is the correct unit used to report the measurement of

$A$ nucleus decays by ${\beta ^ + }$ emission followed by a gamma emission. If the atomic and mass numbers of the parent nucleus are $Z$ and $A$ respectively, the corresponding numbers for the daughter nucleus are respectively.

Vedclass Test Series

Exam Paper Generator

Online Exam Module