In the given radioactive decay if the mass number and atomic number of D2 are 176 and 71 respectively.What is the mass number and atomic number of D? D----->D1------->D2 Bita alpha particle

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In the given radioactive decay if the mass number and atomic number of D2 are 176 and 71 respectively.What is the mass number and atomic number of D? D----->D1------->D2 Bita alpha particle

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Answer 1 · 2018-03-01T06:26:06

Well,though not clearly mentioned,considering,the reaction to be,

$D - - - - - \to D_{1} + β - - - - - \to D_{2} + α$ $D ------> D_1 + beta ------> D_2+alpha$

Given, mass number of $D_{2}$ $D_2$ is $176$ $176$ , now as $D_{2}$ $D_2$ has been formed from $D_{1}$ $D_1$ due to $α$ $alpha$ emission,mass number of $D_{1}$ $D_1$ must be $(176 + 4) - 180$ $(176+4)-180$ and similarly atomic number should be $71 + 2 = 73$ $71+2=73$ (as, alpha particle has mass number $4$ $4$ and atomic number $2$ $2$ )

So, we can complete the $2 n d$ $2nd$ half of the reaction,

$_{73}^{180} D_{2} - - - - - - \to_{71}^{176} D_{3} +_{2}^{4} H e$ $""_73^180 D_2 -------> ""_71^176 D_3 + ""_2^4 He$

Similarly, we can say, there will be no change in mass number of $D$ $D$ as,beta particle has mass number of $0$ $0$ as beta emission is expressed as, $_{- 1}^{0} β$ $""_(-1)^ 0 beta$ ,So its atomic number must decrease by $1$ $1$ i.e $72$ $72$

So,the whole reaction becomes,

$_{72}^{180} D - - - \to_{73}^{180} D_{2} +_{- 1}^{0} β - - - - - - \to_{71}^{176} D_{3} +_{2}^{4} H e$ $""_72^180 D ---->""_73^180 D_2 + ""_-1^0 beta -------> ""_71^176 D_3 + ""_2^4 He$

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In the given radioactive decay if the mass number and atomic number of D2 are 176 and 71 respectively.What is the mass number and atomic number of D? D----->D1------->D2 Bita alpha particle

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