In successive emission of alpha - and beta - particles- the number of alpha - and beta - particles that should be emitted the conversion of -92-U-238- to -82-Pb-206- are7 alpha- 5 beta6 alpha- 4 beta4 alpha- 3 beta8 alpha- 6 beta

Question

Toppr Admin · Accepted Answer

The correct option is D 8-x3B1-6-x3B2-92U238-x27F6-82Pb206-m2He4-n-x2212-1e0On comparing both sides238-206-4m-x21D2-m-892-82-2m-x2212-n2m-x2212-n-104-x21D2-n-6-x2234-x3B1- - particles emitted- m-8-x3B2- - particles emitted- n-6

In successive emission of $α$ - and $β$ - particles, the number of $α$ - and $β$ - particles that should be emitted for the conversion of $_{92} U^{238}$ to $_{82} {P b}^{206}$ are
$7 α, 5 β$
$6 α, 4 β$
$4 α, 3 β$
$8 α, 6 β$

The correct option is D $8 α, 6 β$
$_{92} U^{238} ⟶_{82} {P b}^{206} + m_{2} {H e}^{4} + n_{- 1} e^{0}$
On comparing both sides
$238 = 206 + 4 m$
$\Rightarrow m = 8$
$92 = 82 + 2 m - n$
$2 m - n = 10$ 4
$\Rightarrow n = 6$
$∴ α$ - particles emitted, $m = 8$
$β$ - particles emitted, $n = 6$

In successive emission of α - and β - particles, the number of α - and β - particles that should be emitted for the conversion of 92U238 to 82Pb206 are7α,5β6α,4β4α,3β8α,6β

The correct option is D 8α,6β92U238⟶82Pb206+m2He4+n−1e0On comparing both sides238=206+4m⇒m=892=82+2m−n2m−n=104⇒n=6∴α - particles emitted, m=8β - particles emitted, n=6

In successive emission of $α$ - and $β$ - particles, the number of $α$ - and $β$ - particles that should be emitted for the conversion of $_{92} U^{238}$ to $_{82} {P b}^{206}$ are
$7 α, 5 β$
$6 α, 4 β$
$4 α, 3 β$
$8 α, 6 β$

The correct option is D $8 α, 6 β$
$_{92} U^{238} ⟶_{82} {P b}^{206} + m_{2} {H e}^{4} + n_{- 1} e^{0}$
On comparing both sides
$238 = 206 + 4 m$
$\Rightarrow m = 8$
$92 = 82 + 2 m - n$
$2 m - n = 10$ 4
$\Rightarrow n = 6$
$∴ α$ - particles emitted, $m = 8$
$β$ - particles emitted, $n = 6$