(a) the number that goes into the elevated box (superscript) in the given nuclear reaction. (b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 43, Problem 1Q

To determine

To find:

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction.

(b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) The number that goes into the elevated box (superscript) in the given nuclear reaction = 42

(b) The number that goes into the descended box (value of Z) in the given nuclear reaction = 101

Explanation of Solution

Given:

i. A nuclear reaction ²³⁵U + n ¹³²Sn + + 3n .

ii. Refer to Appendix F and G of the textbook to know about A and Z of elements.

Concept

In any nuclear reaction, A and Z has to be conserved.

In the given reaction, balance A and Z on both sides of the reaction, because these numbers of conserved.

Let us say that the unknown nucleus ‘unknown’ has A=X and Z = Y

²³⁵U + n ¹³²Sn + unknownYX + 3n

(a)

On the left-hand side:

A = 235+1 = 236 ………………………………………………(1)

Z = 92 +0 = 92 …………………………………………………(2)

On the right-hand side:

A = 132 + X+ 3 = 135 + X ……………………………………..(3)

Z = 50 + Y + 0 = 50+ Y ………………………………………..(4)

Comparing eqn (1) and eqn (3) we can calculate that

X = 236 – 135 = 101

(b)

Comparing eqn (2) and (4) we can calculate that

Y = 91 – 50 = 42

From Appendix F and G, we can say that the ‘unknown’ element is Molybdenum Mo.

We can now write the complete nuclear reaction as

²³⁵U + n ¹³²Sn + Mo42101 + 3n

Conclusion

We can complete a given nuclear reaction by calculating A and Z of elements on both the sides of the reaction.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Sketch the harmonic on graphing paper.

Exercise 1: (a) Using the explicit formulae derived in the lectures for the (2j+1) × (2j + 1) repre- sentation matrices Dm'm, (J/h), derive the 3 × 3 matrices corresponding to the case j = 1. (b) Verify that they satisfy the so(3) Lie algebra commutation relation: [D(Î₁/ħ), D(Î₂/h)]m'm₁ = iƊm'm² (Ĵ3/h). (c) Prove the identity 3 Dm'm,(Î²) = Σ (D(Î¡)D(Î¡))m'¡m; · i=1

Sketch the harmonic.

Answer 2

Question

Chapter 43, Problem 1Q

To determine

To find:

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction.

(b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) The number that goes into the elevated box (superscript) in the given nuclear reaction = 42

(b) The number that goes into the descended box (value of Z) in the given nuclear reaction = 101

Explanation of Solution

Given:

i. A nuclear reaction ²³⁵U + n ¹³²Sn + + 3n .

ii. Refer to Appendix F and G of the textbook to know about A and Z of elements.

Concept

In any nuclear reaction, A and Z has to be conserved.

In the given reaction, balance A and Z on both sides of the reaction, because these numbers of conserved.

Let us say that the unknown nucleus ‘unknown’ has A=X and Z = Y

²³⁵U + n ¹³²Sn + unknownYX + 3n

(a)

On the left-hand side:

A = 235+1 = 236 ………………………………………………(1)

Z = 92 +0 = 92 …………………………………………………(2)

On the right-hand side:

A = 132 + X+ 3 = 135 + X ……………………………………..(3)

Z = 50 + Y + 0 = 50+ Y ………………………………………..(4)

Comparing eqn (1) and eqn (3) we can calculate that

X = 236 – 135 = 101

(b)

Comparing eqn (2) and (4) we can calculate that

Y = 91 – 50 = 42

From Appendix F and G, we can say that the ‘unknown’ element is Molybdenum Mo.

We can now write the complete nuclear reaction as

²³⁵U + n ¹³²Sn + Mo42101 + 3n

Conclusion

We can complete a given nuclear reaction by calculating A and Z of elements on both the sides of the reaction.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 43, Problem 1Q

To determine

To find:

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction.

(b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) The number that goes into the elevated box (superscript) in the given nuclear reaction = 42

(b) The number that goes into the descended box (value of Z) in the given nuclear reaction = 101

Explanation of Solution

Given:

i. A nuclear reaction ²³⁵U + n ¹³²Sn + + 3n .

ii. Refer to Appendix F and G of the textbook to know about A and Z of elements.

Concept

In any nuclear reaction, A and Z has to be conserved.

In the given reaction, balance A and Z on both sides of the reaction, because these numbers of conserved.

Let us say that the unknown nucleus ‘unknown’ has A=X and Z = Y

²³⁵U + n ¹³²Sn + unknownYX + 3n

(a)

On the left-hand side:

A = 235+1 = 236 ………………………………………………(1)

Z = 92 +0 = 92 …………………………………………………(2)

On the right-hand side:

A = 132 + X+ 3 = 135 + X ……………………………………..(3)

Z = 50 + Y + 0 = 50+ Y ………………………………………..(4)

Comparing eqn (1) and eqn (3) we can calculate that

X = 236 – 135 = 101

(b)

Comparing eqn (2) and (4) we can calculate that

Y = 91 – 50 = 42

From Appendix F and G, we can say that the ‘unknown’ element is Molybdenum Mo.

We can now write the complete nuclear reaction as

²³⁵U + n ¹³²Sn + Mo42101 + 3n

Conclusion

We can complete a given nuclear reaction by calculating A and Z of elements on both the sides of the reaction.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 43, Problem 1Q

To determine

To find:

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction.

(b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) The number that goes into the elevated box (superscript) in the given nuclear reaction = 42

(b) The number that goes into the descended box (value of Z) in the given nuclear reaction = 101

Explanation of Solution

Given:

i. A nuclear reaction ²³⁵U + n ¹³²Sn + + 3n .

ii. Refer to Appendix F and G of the textbook to know about A and Z of elements.

Concept

In any nuclear reaction, A and Z has to be conserved.

In the given reaction, balance A and Z on both sides of the reaction, because these numbers of conserved.

Let us say that the unknown nucleus ‘unknown’ has A=X and Z = Y

²³⁵U + n ¹³²Sn + unknownYX + 3n

(a)

On the left-hand side:

A = 235+1 = 236 ………………………………………………(1)

Z = 92 +0 = 92 …………………………………………………(2)

On the right-hand side:

A = 132 + X+ 3 = 135 + X ……………………………………..(3)

Z = 50 + Y + 0 = 50+ Y ………………………………………..(4)

Comparing eqn (1) and eqn (3) we can calculate that

X = 236 – 135 = 101

(b)

Comparing eqn (2) and (4) we can calculate that

Y = 91 – 50 = 42

From Appendix F and G, we can say that the ‘unknown’ element is Molybdenum Mo.

We can now write the complete nuclear reaction as

²³⁵U + n ¹³²Sn + Mo42101 + 3n

Conclusion

We can complete a given nuclear reaction by calculating A and Z of elements on both the sides of the reaction.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 43, Problem 1Q

To determine

To find:

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction.

(b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) The number that goes into the elevated box (superscript) in the given nuclear reaction = 42

(b) The number that goes into the descended box (value of Z) in the given nuclear reaction = 101

Explanation of Solution

Given:

i. A nuclear reaction ²³⁵U + n ¹³²Sn + + 3n .

ii. Refer to Appendix F and G of the textbook to know about A and Z of elements.

Concept

In any nuclear reaction, A and Z has to be conserved.

In the given reaction, balance A and Z on both sides of the reaction, because these numbers of conserved.

Let us say that the unknown nucleus ‘unknown’ has A=X and Z = Y

²³⁵U + n ¹³²Sn + unknownYX + 3n

(a)

On the left-hand side:

A = 235+1 = 236 ………………………………………………(1)

Z = 92 +0 = 92 …………………………………………………(2)

On the right-hand side:

A = 132 + X+ 3 = 135 + X ……………………………………..(3)

Z = 50 + Y + 0 = 50+ Y ………………………………………..(4)

Comparing eqn (1) and eqn (3) we can calculate that

X = 236 – 135 = 101

(b)

Comparing eqn (2) and (4) we can calculate that

Y = 91 – 50 = 42

From Appendix F and G, we can say that the ‘unknown’ element is Molybdenum Mo.

We can now write the complete nuclear reaction as

²³⁵U + n ¹³²Sn + Mo42101 + 3n

Conclusion

We can complete a given nuclear reaction by calculating A and Z of elements on both the sides of the reaction.

Answer 6

Chapter 43, Problem 1Q

To determine

To find:

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction.

(b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) The number that goes into the elevated box (superscript) in the given nuclear reaction = 42

(b) The number that goes into the descended box (value of Z) in the given nuclear reaction = 101

Explanation of Solution

Given:

i. A nuclear reaction ²³⁵U + n ¹³²Sn + + 3n .

ii. Refer to Appendix F and G of the textbook to know about A and Z of elements.

Concept

In any nuclear reaction, A and Z has to be conserved.

In the given reaction, balance A and Z on both sides of the reaction, because these numbers of conserved.

Let us say that the unknown nucleus ‘unknown’ has A=X and Z = Y

²³⁵U + n ¹³²Sn + unknownYX + 3n

(a)

On the left-hand side:

A = 235+1 = 236 ………………………………………………(1)

Z = 92 +0 = 92 …………………………………………………(2)

On the right-hand side:

A = 132 + X+ 3 = 135 + X ……………………………………..(3)

Z = 50 + Y + 0 = 50+ Y ………………………………………..(4)

Comparing eqn (1) and eqn (3) we can calculate that

X = 236 – 135 = 101

(b)

Comparing eqn (2) and (4) we can calculate that

Y = 91 – 50 = 42

From Appendix F and G, we can say that the ‘unknown’ element is Molybdenum Mo.

We can now write the complete nuclear reaction as

²³⁵U + n ¹³²Sn + Mo42101 + 3n

Conclusion

We can complete a given nuclear reaction by calculating A and Z of elements on both the sides of the reaction.

Answer 7

Chapter 43, Problem 1Q

To determine

To find:

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction.

(b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Answer 8

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) The number that goes into the elevated box (superscript) in the given nuclear reaction = 42

(b) The number that goes into the descended box (value of Z) in the given nuclear reaction = 101

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

i. A nuclear reaction ²³⁵U + n ¹³²Sn + + 3n .

ii. Refer to Appendix F and G of the textbook to know about A and Z of elements.

Concept

In any nuclear reaction, A and Z has to be conserved.

In the given reaction, balance A and Z on both sides of the reaction, because these numbers of conserved.

Let us say that the unknown nucleus ‘unknown’ has A=X and Z = Y

²³⁵U + n ¹³²Sn + unknownYX + 3n

(a)

On the left-hand side:

A = 235+1 = 236 ………………………………………………(1)

Z = 92 +0 = 92 …………………………………………………(2)

On the right-hand side:

A = 132 + X+ 3 = 135 + X ……………………………………..(3)

Z = 50 + Y + 0 = 50+ Y ………………………………………..(4)

Comparing eqn (1) and eqn (3) we can calculate that

X = 236 – 135 = 101

(b)

Comparing eqn (2) and (4) we can calculate that

Y = 91 – 50 = 42

From Appendix F and G, we can say that the ‘unknown’ element is Molybdenum Mo.

We can now write the complete nuclear reaction as

²³⁵U + n ¹³²Sn + Mo42101 + 3n

Conclusion

We can complete a given nuclear reaction by calculating A and Z of elements on both the sides of the reaction.

Answer 12

Ch. 43 - Prob. 1Q Ch. 43 - Prob. 2Q Ch. 43 - Prob. 3Q Ch. 43 - Prob. 4Q Ch. 43 - Prob. 5Q Ch. 43 - Prob. 6Q Ch. 43 - Prob. 7Q Ch. 43 - Which of these elements is not cooked up by...Ch. 43 - Prob. 9Q Ch. 43 - Prob. 10Q

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction. (b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Concept explainers

To find:

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction.

(b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Answer to Problem 1Q

Explanation of Solution

Want to see more full solutions like this?

Chapter 43 Solutions

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction. (b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Concept explainers

To find: (a) the number that goes into the elevated box (superscript) in the given nuclear reaction. (b) the number that goes into the descended box (value of Z) in the given nuclear reaction.

Answer to Problem 1Q

Explanation of Solution

Want to see more full solutions like this?

Chapter 43 Solutions

To find:

(a) the number that goes into the elevated box (superscript) in the given nuclear reaction.

(b) the number that goes into the descended box (value of Z) in the given nuclear reaction.