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Question

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\[_{\text{4}}^{\text{9}}\text{Be + X }\to _{\text{4}}^{\text{8}}\text{Be + Y}\],

X and Y are:

(This question has multiple correct options)

A) $\text{ }\!\!\gamma\!\!\text{ ,n}$

B) $\text{P,D}$

C) $\text{n,D}$

D) $\gamma \text{,P}$

Answer
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- Radioactive elements undergo the process of decay under the natural condition to form a new element with the emission of radiation. Nuclear reactions involve the decay of radioactive elements to form a new nucleus.

- Transmutation reactions involve the conversion of one chemical element into another chemical element. The transmutation process can also occur naturally.

We are provided with the following nuclear reaction,

\[_{\text{4}}^{\text{9}}\text{Be + X }\to _{\text{4}}^{\text{8}}\text{Be + Y}\]

We have to find the X and Y in the nuclear transmutation reaction.

We are well familiar that chemistry equations can be balanced using a simple rule. In the transmutation reaction, the mass number of the reactant and product are conserved during the nuclear reaction.

If there is more than one reactant or product then the sum of the mass number should remain the same in the progress of the reaction. The electric charge is also conserved in transmutation reactions.

Let's find out the suitable values for X and Y.

A) $\text{ }\!\!\gamma\!\!\text{ ,n}$

Let $\text{X =}\gamma \text{ , Y=n}$

The nuclear reaction becomes, \[_{4}^{9}\text{Be+}_{0}^{0}\gamma \to _{4}^{9}\text{Be}+_{0}^{1}n\]

The $\gamma $ radiations are depicted as $_{\text{0}}^{\text{0}}\text{ }\!\!\gamma\!\!\text{ }$and the neutron is as$_{\text{0}}^{\text{1}}\text{n}$.

On putting X and Y we get,

\[_{4}^{9}\text{Be+}_{0}^{0}\gamma \to _{4}^{9}\text{Be}{{+}^{0}}n\]

Since the mass number on the reactant side and product side are the same which is 9 and the charges are conserved for the reaction. Thus option (A) is the correct option.

B) Here $\text{X=P , Y=D}$ where P is the proton having a relative charge $_{\text{1}}^{\text{1}}\text{H}$ and D is the deuterium $_{\text{1}}^{\text{2}}\text{H}$Lets substitute in reaction,

\[_{4}^{9}\text{Be+}_{1}^{1}\text{H}\to _{4}^{8}\text{Be}+_{1}^{2}\text{D}\]

Here, the mass number on the reactant is 10 and equal to the mass number on the product. Thus $\text{(p,d)}$is a transmutation reaction. Option B) is the correct option.

C) Here $\text{X=n , Y=D}$ where P is the neutron having a relative charge $^{0}n$ and D is the deuterium $_{\text{1}}^{\text{2}}\text{H}$

Let’s substitute in reaction,

\[_{4}^{9}\text{Be+}_{0}^{1}n\to _{4}^{8}\text{Be}+_{1}^{2}\text{H}\]

Here, the mass number on the reactant is equal to the mass number on the product but the atomic number is not equal. Thus this does not satisfy the condition for transmutation.

Option C) cannot be true.

D) Here $\text{X=}\gamma \text{ , Y=P}$ where P is the neutron having a relative charge $^{0}n$ and D is the deuterium $_{\text{1}}^{\text{2}}\text{H}$Lets substitute in reaction,

\[_{4}^{9}\text{Be+}_{0}^{0}\gamma \to _{4}^{8}\text{Be}+_{1}^{1}\text{H}\]

Here, the mass number on the reactant is equal to the mass number on the product but the atomic number is not equal. Thus this does not satisfy the condition for transmutation.

Option D) cannot be true.

The transmutation reactions can be studied when the reactions are balanced. The equations can be balanced by simple rules. The mass number for the reactant and the product must be conserved. The electric charge must be conserved for transmutation reaction.

This is the nuclear reaction of nuclei with elementary particles which results in the emission of some particles. These particles can be an alpha particle, neutrons, beta particle, gamma particles or proton, deuteron, etc. The unstable radioactive nuclei convert itself into stable nuclei by transmutation reaction.

For example,

$\begin{align}

& _{\text{7}}^{\text{14}}\text{N+}_{\text{2}}^{\text{4}}\text{He}\to _{\text{8}}^{\text{17}}\text{O+}_{\text{1}}^{\text{1}}\text{H} \\

& _{\text{26}}^{\text{58}}\text{Fe+}_{\text{0}}^{\text{1}}\text{n}\to _{\text{26}}^{\text{59}}\text{Fe}\to _{\text{27}}^{\text{59}}\text{Co+}_{\text{-1}}^{\text{0}}\text{e} \\

\end{align}$.