EBK CHEMISTRY
EBK CHEMISTRY
4th Edition
ISBN: 8220102797864
Author: Burdge
Publisher: YUZU
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Chapter 20, Problem 38QP
Interpretation Introduction

Interpretation:

The balanced nuclear equations for the given reactions are to be written and the component X is to be identified.

Concept introduction:

In a balanced radioactive decay

Reactant mass number = Product mass number

And

Reactant atomic number = Product atomic number

In the balanced radioactive decay, the subscript present on the individual particles represents the atomic number, while the superscript represents the atomic mass.

While writing the abbreviated form of any radioactive reaction, the reactant is written first followed by parenthesis, which contains the bombarded particle first and then the emitted particle. After this the product is written.

Expert Solution & Answer
Check Mark

Answer to Problem 38QP

Solution:

(a) 157N +11 126C + 42α and X is N715.

(b) 2713Al +12 2512Mg + 42α and X is 2512Mg.

(c) 5525Mn +01 5625Mn + γ and X is 5625Mn.

Explanation of Solution

a) X(p,α)126C.

It is known that while writing the abbreviated form, the reactant is written first.

Here, the reactant is X.

Then, it is followed by the parentheses where the bombarded particle is written first, followed by the emitted particles. These are separated by commas.

Here, the bombarded particle is 11p while the emitted particle is 42α.

At the end, the parenthesis is closed and the product is written.

Here, the product is 126C.

Thus, the equation can be written as

X +11 126C + 42α

Now, apply the balancing rules as

Reactant mass number = Product mass number

And

Reactant atomic number = Product atomic number

On the left side of the reaction, the sum of the atomic numbers is given by the expression as follows: 1 + atomic number  of X

On the right-hand side of the reaction, the sum of the atomic numbers is calculated as follows:

6+ 2 = 8

Thus, the atomic number of X is calculated as follows:

(8)(1)=7

Similarly,

On the left side of the reaction, the sum of the mass numbers is given by the expression as follows:

1 + mass number of X

On the right-hand side of the reaction, the sum of the mass numbers is as follows:

12 + 4 = 16

Thus, the mass number of X is calculated as follows:

(16)(1)=15

So, X will be 157N.

Hence, the balanced reaction is given as follows:

157N +11 126C + 42α

b) 2713Al(d,α)X

It is known that while writing the abbreviated form, the reactant is written first.

Here, the reactant is 2713Al

Then, it is followed by the parenthesis where the bombarded particle is written first followed by the emitted particles. These are separated by commas.

Here, the bombarded particle is 21d while the emitted particle is 42α.

At the end, the parenthesis is closed and the product is written.

Here, the product is X.

Thus, the equation can be written as follows:

2713Al +12 X + 42α

Now, apply the balancing rules as follows:

Reactant mass number = Product mass number

And

Reactant atomic number = Product atomic number

On the left side of the reaction, the sum of the atomic numbers is given as follows:

13 + 1 = 14

On the right hand side of the reaction, the sum of the atomic numbers is

atomic number of X+ 2 

Thus, the atomic number of X is calculated as follows:

(14)(2)=12

Similarly,

On the left side of the reaction, the sum of the mass numbers is calculated as follows:

27 + 2 = 29

On the right hand side of the reaction, the sum of the mass numbers is

Atomic mass of X + 4

Thus, the mass number of X is calculated as follows:

(29)(4)=25

So, X will be 2512Mg.

Hence, the balanced reaction is given as follows:

2713Al +12 2512Mg + 42α

c) 5525Mn(n,γ)X

It is known that while writing the abbreviated form, the reactant is written first.

Here, the reactant is 5525Mn

Then, it is followed by the parenthesis where the bombarded particle is written first followed by the emitted particles. These are separated by commas.

Here, the bombarded particle is 10n, while the emitted particle is γ.

At the end, the parenthesis is closed and the product is written.

Here, the product is X.

Thus, the equation can be written as follows:

5525Mn +01 X + γ

Now, apply the balancing rules as follows:

Reactant mass number = Product mass number

And

Reactant atomic number = Product atomic number

On the left side of the reaction, the sum of the atomic numbers is given as follows:

25 + 0 = 25

On the right hand side of the reaction, the sum of the atomic numbers is

Atomic number of X+ 0

Thus, the atomic number of X is calculated as follows:

(25)(0)=25

Similarly,

On the left side of the reaction, the sum of the mass numbers is calculated as follows:

55 + 1 = 56

On the right hand side of the reaction, the sum of the mass numbers is

Atomic mass of X + 0

Thus, the mass number of X is calculated as follows:

(56)(0)=56

So, X will be 5625Mn

Hence, the balanced reaction is given as follows:

5525Mn +01 5625Mn + γ

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Chapter 20 Solutions

EBK CHEMISTRY

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