Q24.54 P - Chemistry: Molecular Nature Of Matter And Change

Question

Name the unidentified species, and write each transmutation process in shorthand notation:

 (a) gamma irradiation of a nuclide yields a proton, a neutron, and 29Si; 

(b) bombardment of 252Cf with 10B yields five neutrons and a nuclide; 

(c) bombardment of 238U with a particle yields three neutrons and 239Pu.

 

Step-by-Step Solution

Answer

$\begin{array}{l}{\text{a)}}_{\text{15}}^{\text{31}}{\text{P(γ,p and n)}}_{\text{14}}^{\text{29}}\text{Si}\\ {\text{b)}}_{\text{98}}^{\text{252}}\text{Cf}{{(}_{\text{5}}^{\text{10}}\text{ B,5n})}_{\text{103}}^{\text{257}}\text{Lr}\\ {\text{c)}}_{\text{92}}^{\text{238}}{\text{U(α,3n)}}_{\text{94}}^{\text{239}}\text{Pu}\end{array}$

Step 1: Subpart (a)

Mass number (the sum of protons and neutrons) is represented with A.

Atomic number (the number of protons) is represented with Z.

Let X be the unknown element or particle.

(a) gamma irradiation of a nuclide yields a proton, a neutron, and  - Atomic number of Si is $14$ 

- Gamma ray has neither charge nor mass

- Proton: ${}_{\text{1}}^{\text{1}}\text{P}$ 

- Neutron: ${}_{\text{0}}^{\text{1}}\text{n}$ 

 _{${\text{A}}_{\text{Z}}^{\text{A}}\text{+γ}{\to }_{\text{1}}^{\text{1}}{\text{p+}}_{\text{0}}^{\text{1}}{\text{n+}}_{\text{14}}^{\text{29}}\text{Si}$}

First, let us calculate the values of A and Z

_{$\begin{array}{l}\text{A=1+1+29}\\ \text{ A=31Z=1+0+14Z=15}\end{array}$} 

Hence The element with atomic number 15 is phosphorus, so the reactant is  _{${}_{15}^{31}\text{P}$}, and the balanced equation is

 _{${}_{\text{15}}^{\text{31}}\text{P+γ}{\to }_{\text{1}}^{\text{1}}{\text{p+}}_{\text{0}}^{\text{1}}{\text{n+}}_{\text{14}}^{\text{29}}\text{Si}$}

- Shorthand notation

 _{${}_{\text{15}}^{\text{31}}{\text{P(γ,p and n)}}_{\text{14}}^{\text{29}}\text{Si}$}

Step 2: Subpart (b)

 Bombardment of ${}^{\text{252}}\text{Cf}$ with   ${}^{10} \text{B}$ yields five neutrons and a nuclide

- Atomic number of $\text{Cf}$  is$98$  

- Atomic number of B is 5

- Neutron: ${}_{\text{0}}^{\text{1}}\text{n}$ 

 _{${}_{\text{98}}^{\text{252}}{\text{Cf+}}_{\text{5}}^{\text{10}}\text{ B}\to {\text{5}}_{\text{0}}^{\text{1}}{\text{n+}}_{\text{Z}}^{\text{A}}\text{X}$}

First, let us calculate the values of A and Z

 _{$\begin{array}{c}\text{252+10 =5×1+A }\\ \text{A=257}\\ \text{98+5=5×0+Z }\\ \text{Z=103}\end{array}$}

Hence, the element with atomic number 103 is lawrencium, so the product is${}_{103}^{257}\text{Lr}$  , and the balanced equation is

_{${}_{98}^{252}\text{Cf}{+}_5^{10} \text{B}\to 5_0^1\text{n}{+}_{103}^{257}\text{Lr}$} 

- Shorthand notation

_{${}_{98}^{252}\text{Cf}{\left(\frac{{\displaystyle 10}}{{\displaystyle 5}} \text{B},5\text{n}\right)}_{103}^{257}\text{Lr}$} 

Step 3: Subpart (c)

Bombardment of  with a particle yields three neutrons and ${}^{239}\text{Pu}$ - Atomic number of U is  _$92$.

- Atomic number of Pu is 94

- Neutron: ${}_{\text{0}}^{\text{1}}\text{n}$ 

 _{${}_{92}^{238}\text{U}{+}_{\text{Z}}^{\text{A}}\text{X}\to 3_0^1\text{n}{+}_{94}^{239}\text{Pu}$}

First, let us calculate the values of A and Z

 _{$\begin{array}{c}\text{238+A =3×1+239}\\ \text{ A =4}\\ \text{ 92+Z =3×0+94}\\ \text{ Z=2}\end{array}$}

Hence, the particle with mass number 4 and charge 2 is alpha particle  ,${(}_2^4\text{He})$ and the balanced equation is

 _{${}_{92}^{238}\text{U}{+}_2^4\text{He}\to 3_0^1\text{n}{+}_{94}^{239}\text{Pu}$}

- Shorthand notation

 _{${}_{\text{92}}^{\text{238}}{\text{U(α,3n)}}_{\text{94}}^{\text{239}}\text{Pu}$}