Question

Complete each of the following nuclear equations and describe the type of radiation:

(a)${}_{13}{}^{28}\mathrm{Al}\to ?+{}_{-1}{}^0\mathrm{e}$

(b)${}_{73}{}^{180}Ta\to {}_{73}{}^{180}\mathrm{Ta}+?$

(c)${}_{29}{}^{66}\mathrm{Cu}\to {}_{30}{}^{66}\mathrm{Zn}+?$

(d)$?\to {}_{90}{}^{234}\mathrm{Th}+{}_2{}^4\mathrm{He}$

(e)${}_{80}{}^{188}\mathrm{Hg}\to ?+{}_1{}^0\mathrm{e}$

Short answer

(a) The complete balanced nuclear equation is${}_{13}{}^{28}\mathrm{Al}\to {}_{14}{}^{28}\mathrm{Si}+{}_{-1}{}^0\mathrm{e}$.

It is beta decay.

(b) The complete balanced nuclear equation is${}_{73}{}^{180}\mathrm{Ta}\to {}_{73}{}^{180}\mathrm{Ta}+{}_0{}^0\mathrm{e}$. It is gamma emission.

(c) The complete balanced nuclear equation is${}_{29}{}^{66}\mathrm{Cu}\to {}_{30}{}^{66}\mathrm{Zn}+{}_{-1}{}^0\mathrm{e}$. It is beta decay.

(d) The complete balanced nuclear equation is${}_{92}{}^{238}\mathrm{U}\to {}_{90}{}^{234}\mathrm{Th}+{}_2{}^4\mathrm{He}$. It is alpha decay.

(e) The complete balanced nuclear equation is${}_{80}{}^{188}\mathrm{Hg}\to {}_{79}{}^{188}Au+{}_1{}^0\mathrm{e}$. It is positron emission.

Step-by-step solution

Part (a) Step 1: Given Information

We are given an incomplete equation and we have to complete it and also describe its type.

Part (a) Step 2: Explanation

To complete the equation,

${}_{13}{}^{28}\mathrm{Al}\to ?+{}_{-1}{}^0\mathrm{e}$,

equate the mass number,

We get$28=?+0$,

So, the mass number is$28$.

Now, equating the atomic number,

We get$13=?-1$,

So, the atomic number is$14$.

So, the element is silicon.

Hence, the equation is ${}_{13}{}^{28}\mathrm{Al}\to {}_{14}{}^{28}\mathrm{Si}+{}_{-1}{}^0\mathrm{e}$and as the atomic number is increasing by one so, it is beta decay.

Part (b) Step 1: Given Information

We are given an incomplete equation and we have to complete it and also describe its type.

Part (b) Step 2: Explanation

To complete the equation,

${}_{73}{}^{180}\mathrm{Ta}\to {}_{73}{}^{180}\mathrm{Ta}+?$,

equate the mass number,

We get$180=180+?$,

So, the mass number is$0$.

Now, equating the atomic number ,

We get$73=73+0$,

So, the atomic number is$0$.

So, the element is e.

Hence, the equation isrole="math" localid="1653726718211" ${}_{73}{}^{180}\mathrm{Ta}\to {}_{73}{}^{180}\mathrm{Ta}+{}_0{}^0\mathrm{e}$ and as the atomic number is the same, it is gamma emission.

Part (c) Step 1: Given Information

We are given an incomplete equation and we have to complete it and also describe its type.

Part (c) Step 2: Explanation

To complete the equation,

${}_{29}{}^{66}\mathrm{Cu}\to {}_{30}{}^{66}\mathrm{Zn}+?$,

equate the mass number,

We get$66=66+0$,

So, the mass number is$0$.

Now, equating the atomic number ,

We getrole="math" localid="1653725772295" $29=30+?$,

So, the atomic number is$-1$

So, the element is e.

Hence, the equation isrole="math" localid="1653726777789" ${}_{29}{}^{66}\mathrm{Cu}\to {}_{30}{}^{66}\mathrm{Zn}+{}_{-1}{}^0\mathrm{e}$ and as the atomic number is increasing by one, it is beta decay.

Part (d) Step 1: Given Information

We are given an incomplete equation and we have to complete it and also describe its type.

Part (d) Step 2: Explanation

To complete the equation,

$?\to {}_{90}{}^{234}\mathrm{Th}+{}_2{}^4\mathrm{He}$,

equate the mass number,

We get$?=234+4$,

So, the mass number isrole="math" localid="1653726147905" $238$.

Now, equating the atomic number ,

We get$?=90+2$,

So, the atomic number is$92$.

So, the element is uranium.

Hence, the equation isrole="math" localid="1653726800946" ${}_{92}{}^{238}\mathrm{U}\to {}_{90}{}^{234}\mathrm{Th}+{}_2{}^4\mathrm{He}$and as the atomic number is decreasing by two, it is alpha decay.

Part (e) Step 1: Given Information

We are given an incomplete equation and we have to complete it and also describe its type.

Part (e) Step 2: Explanation

To complete the equation,

${}_{80}{}^{188}\mathrm{Hg}\to ?+{}_1{}^0\mathrm{e}$,

equate the mass number,

We get$188=?+0$,

So, the mass number is$188$.

Now, equating the atomic number ,

We get$80=?+1$,

So, the atomic number is$79$.

So, the element is gold .

Hence, the equation isrole="math" localid="1653726863968" ${}_{80}{}^{188}\mathrm{Hg}\to {}_{79}{}^{188}Au+{}_1{}^0\mathrm{e}$ and as the atomic number is decreasing by one, it is positron emission.