In the successive emission of alpha- and beta | vedclass

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Question

(A) The nuclear reaction is: ${}^{238}_{92}U \longrightarrow {}^{206}_{82}Pb + m({}^{4}_{2}He) + n({}^{0}_{-1}e)$.Comparing the mass numbers on both s

Vedclass · Accepted Answer

$8 \alpha, 6 \beta$

Vedclass · Answer

(A) The nuclear reaction is: ${}^{238}_{92}U \longrightarrow {}^{206}_{82}Pb + m({}^{4}_{2}He) + n({}^{0}_{-1}e)$.Comparing the mass numbers on both sides: $238 = 206 + 4m$ $\Rightarrow 4m = 32$ $\Rightarrow m = 8$.Comparing the atomic numbers on both sides: $92 = 82 + 2m - n$.Substituting $m = 8$: $92 = 82 + 2(8) - n$ $\Rightarrow 92 = 82 + 16 - n$ $\Rightarrow 92 = 98 - n$ $\Rightarrow n = 6$.Therefore,$8$ $\alpha$-particles and $6$ $\beta$-particles are emitted.

List-$I$ (Nuclear reactor component)	List-$II$ (Used substance)
$1$. Moderator	$(a)$ Uranium
$2$. Control rods	$(b)$ Graphite
$3$. Fuel rods	$(c)$ Boron
$4$. Coolant	$(d)$ Lead
	$(e)$ Sodium

In the successive emission of $\alpha$- and $\beta$-particles,the number of $\alpha$- and $\beta$-particles that should be emitted for the conversion of ${}^{238}_{92}U$ to ${}^{206}_{82}Pb$ are:

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Match List-$I$ and List-$II$ and choose the correct option using the codes given below:

List-$I$ (Nuclear reactor component) List-$II$ (Used substance)

$1$. Moderator $(a)$ Uranium

$2$. Control rods $(b)$ Graphite

$3$. Fuel rods $(c)$ Boron

$4$. Coolant $(d)$ Lead

$(e)$ Sodium

Uranium $_{92}U^{235}$ on bombardment with slow neutrons produces:

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