{ }_{92}^{238} U is known to undergo radioact | vedclass

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Question

(D) The decay process is ${ }_{92}^{238} U \rightarrow { }_{82}^{206} Pb + n_{\alpha} { }_{2}^{4} He + n_{\beta} { }_{-1}^{0} e$.Equating mass numbers

Vedclass · Accepted Answer

$1.20$

Vedclass · Answer

(D) The decay process is ${ }_{92}^{238} U ightarrow { }_{82}^{206} Pb + n_{\alpha} { }_{2}^{4} He + n_{\beta} { }_{-1}^{0} e$.Equating mass numbers: $238 = 206 + 4n_{\alpha} \Rightarrow 4n_{\alpha} = 32 \Rightarrow n_{\alpha} = 8$.Initial moles of ${ }_{92}^{238} U = \frac{68 	imes 10^{-6} 	ext{ g}}{238 	ext{ g/mol}} \approx 2.857 	imes 10^{-7} 	ext{ mol}$.In three half-lives,the fraction of nuclei decayed is $1 - (1/2)^3 = 1 - 1/8 = 7/8$.Moles of ${ }_{92}^{238} U$ decayed $= \frac{7}{8} 	imes \frac{68 	imes 10^{-6}}{238} 	ext{ mol}$.Total number of $\alpha$-particles emitted $= (	ext{Moles decayed}) 	imes n_{\alpha} 	imes N_{A}$.$= \frac{7}{8} 	imes \frac{68 	imes 10^{-6}}{238} 	imes 8 	imes 6.022 	imes 10^{23}$.$= 7 	imes \frac{68 	imes 10^{-6}}{238} 	imes 6.022 	imes 10^{23} \approx 1.2044 	imes 10^{18}$.Comparing with $Z 	imes 10^{18}$,we get $Z \approx 1.20$.

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