The atomic radius of xenon having fcc arrangement under given condition has to be determined. Concept Introduction: Relationship of radius with edge length: A fcc unit cell can be represented as given below. Figure 1 In fcc, the corner spheres are in touch with the face centered sphere as shown in the above figure. Hence, the face diagonal A C = 4 R . Consider right angled triangle ACD. ( A C ) 2 = ( A D ) 2 + ( D C ) 2 = a 2 + a 2 = 2 a 2 A C = 2 a = 4 R R = 2 4 a ( o r ) a = 4 R 2

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Answer 1

Question

Chapter 9, Problem 59QRT

Interpretation Introduction

Interpretation:

The atomic radius of xenon having fcc arrangement under given condition has to be determined.

Concept Introduction:

Relationship of radius with edge length:

A fcc unit cell can be represented as given below.

Figure 1

In fcc, the corner spheres are in touch with the face centered sphere as shown in the above figure. Hence, the face diagonal AC= 4R.

Consider right angled triangle ACD.

(AC)2=(AD)2+(DC)2 =a2+a2=2a2AC=2a=4RR=24a (or) a=4R2

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Answer 2

Question

Chapter 9, Problem 59QRT

Interpretation Introduction

Interpretation:

The atomic radius of xenon having fcc arrangement under given condition has to be determined.

Concept Introduction:

Relationship of radius with edge length:

A fcc unit cell can be represented as given below.

Figure 1

In fcc, the corner spheres are in touch with the face centered sphere as shown in the above figure. Hence, the face diagonal AC= 4R.

Consider right angled triangle ACD.

(AC)2=(AD)2+(DC)2 =a2+a2=2a2AC=2a=4RR=24a (or) a=4R2

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Answer 3

Question

Chapter 9, Problem 59QRT

Interpretation Introduction

Interpretation:

The atomic radius of xenon having fcc arrangement under given condition has to be determined.

Concept Introduction:

Relationship of radius with edge length:

A fcc unit cell can be represented as given below.

Figure 1

In fcc, the corner spheres are in touch with the face centered sphere as shown in the above figure. Hence, the face diagonal AC= 4R.

Consider right angled triangle ACD.

(AC)2=(AD)2+(DC)2 =a2+a2=2a2AC=2a=4RR=24a (or) a=4R2

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Answer 4

Question

Chapter 9, Problem 59QRT

Interpretation Introduction

Interpretation:

The atomic radius of xenon having fcc arrangement under given condition has to be determined.

Concept Introduction:

Relationship of radius with edge length:

A fcc unit cell can be represented as given below.

Figure 1

In fcc, the corner spheres are in touch with the face centered sphere as shown in the above figure. Hence, the face diagonal AC= 4R.

Consider right angled triangle ACD.

(AC)2=(AD)2+(DC)2 =a2+a2=2a2AC=2a=4RR=24a (or) a=4R2

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Answer 5

Chapter 9, Problem 59QRT

Interpretation Introduction

Interpretation:

The atomic radius of xenon having fcc arrangement under given condition has to be determined.

Concept Introduction:

Relationship of radius with edge length:

A fcc unit cell can be represented as given below.

Figure 1

In fcc, the corner spheres are in touch with the face centered sphere as shown in the above figure. Hence, the face diagonal AC= 4R.

Consider right angled triangle ACD.

(AC)2=(AD)2+(DC)2 =a2+a2=2a2AC=2a=4RR=24a (or) a=4R2

Answer 6

Chapter 9, Problem 59QRT

Interpretation Introduction

Interpretation:

The atomic radius of xenon having fcc arrangement under given condition has to be determined.

Concept Introduction:

Relationship of radius with edge length:

A fcc unit cell can be represented as given below.

Figure 1

In fcc, the corner spheres are in touch with the face centered sphere as shown in the above figure. Hence, the face diagonal AC= 4R.

Consider right angled triangle ACD.

(AC)2=(AD)2+(DC)2 =a2+a2=2a2AC=2a=4RR=24a (or) a=4R2

Answer 7

Chapter 9, Problem 59QRT

Interpretation Introduction

Interpretation:

The atomic radius of xenon having fcc arrangement under given condition has to be determined.

Concept Introduction:

Relationship of radius with edge length:

A fcc unit cell can be represented as given below.

Figure 1

In fcc, the corner spheres are in touch with the face centered sphere as shown in the above figure. Hence, the face diagonal AC= 4R.

Consider right angled triangle ACD.

(AC)2=(AD)2+(DC)2 =a2+a2=2a2AC=2a=4RR=24a (or) a=4R2

Answer 8

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Answer 11

Ch. 9.1 - Prob. 9.1CE Ch. 9.2 - Prob. 9.2CE Ch. 9.2 - Prob. 9.1PSP Ch. 9.2 - What mass (g) of ethanol, CH3CH2OH(), can be...Ch. 9.3 - Prob. 9.3CE Ch. 9.3 - Prob. 9.4CE Ch. 9.3 - Prob. 9.3PSP Ch. 9.4 - What types of solids are these substances? (a) The...Ch. 9.4 - Prob. 9.5PSP Ch. 9.4 - Prob. 9.5E

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