The probability of finding the electron in the region between r and r + Δ r .

Question

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Answer 1

Question

Chapter 36, Problem 41P

(a)

To determine

The probability of finding the electron in the region between r and r+Δr .

(a)

Expert Solution

Answer to Problem 41P

The probability of finding the electron in the region between r and r+Δr is 9.20×10−4 .

Explanation of Solution

Given:

The value of Δr is 0.02ao .

The value of r is ao .

The value of n is 2 .

The value of l is 0 .

The value of ml is 0 .

Formula used:

The expression for probabilityis given by,

P=P(r)Δr

The expression for P(r) is given by,

P(r)=4πr2|ψ|2

The expression for ψ at n=2 is given by,

ψ=C(2−Zrao)e−Zr/2ao

The expression for C at n=2 is given by,

C=142π(Z a o )3/2

Calculation:

The probability is calculated as,

Solving further as,

P=0.08πaor2(2− Zr a o )2e−Zr/ a o|1 4 2π ( Z a o ) 3/2 |2=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3

For r=ao and Z=1 the above expression is,

P=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3=0.0025ao( a o)2(2− a o a o )2e− a o/ a o( 1 a o )3=9.2×10−4

Conclusion:

Therefore, the probability of finding the electron in the region between r and r+Δr is 9.20×10−4 .

(b)

To determine

The probability of finding the electron in the region between r and r+Δr .

(b)

Expert Solution

Answer to Problem 41P

The probability of finding the electron in the region between r and r+Δr is 0 .

Explanation of Solution

Given:

The value of Δr is 0.02ao .

The value of r is 2ao .

The value of n is 2 .

The value of l is 0 .

The value of ml is 0 .

Formula used:

The expression for probability is given by,

P=P(r)Δr

The expression for P(r) is given by,

P(r)=4πr2|ψ|2

The expression for ψ at n=2 is given by,

ψ=C(2−Zrao)e−Zr/2ao

The expression for C at n=2 is given by,

C=142π(Z a o )3/2

Calculation:

The probability is calculated as,

Solving further as,

P=0.08πaor2(2− Zr a o )2e−Zr/ a o|1 4 2π ( Z a o ) 3/2 |2=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3

For r=2ao and Z=1 the above expression is,

P=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3=0.0025ao(2 a o)2(2− 2 a o a o )2e−2 a o/ a o( 1 a o )3=0

Conclusion:

Therefore, the probability of finding the electron in the region between r and r+Δr is 0 .

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Answer 2

Question

Chapter 36, Problem 41P

(a)

To determine

The probability of finding the electron in the region between r and r+Δr .

(a)

Expert Solution

Answer to Problem 41P

The probability of finding the electron in the region between r and r+Δr is 9.20×10−4 .

Explanation of Solution

Given:

The value of Δr is 0.02ao .

The value of r is ao .

The value of n is 2 .

The value of l is 0 .

The value of ml is 0 .

Formula used:

The expression for probabilityis given by,

P=P(r)Δr

The expression for P(r) is given by,

P(r)=4πr2|ψ|2

The expression for ψ at n=2 is given by,

ψ=C(2−Zrao)e−Zr/2ao

The expression for C at n=2 is given by,

C=142π(Z a o )3/2

Calculation:

The probability is calculated as,

Solving further as,

P=0.08πaor2(2− Zr a o )2e−Zr/ a o|1 4 2π ( Z a o ) 3/2 |2=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3

For r=ao and Z=1 the above expression is,

P=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3=0.0025ao( a o)2(2− a o a o )2e− a o/ a o( 1 a o )3=9.2×10−4

Conclusion:

Therefore, the probability of finding the electron in the region between r and r+Δr is 9.20×10−4 .

(b)

To determine

The probability of finding the electron in the region between r and r+Δr .

(b)

Expert Solution

Answer to Problem 41P

The probability of finding the electron in the region between r and r+Δr is 0 .

Explanation of Solution

Given:

The value of Δr is 0.02ao .

The value of r is 2ao .

The value of n is 2 .

The value of l is 0 .

The value of ml is 0 .

Formula used:

The expression for probability is given by,

P=P(r)Δr

The expression for P(r) is given by,

P(r)=4πr2|ψ|2

The expression for ψ at n=2 is given by,

ψ=C(2−Zrao)e−Zr/2ao

The expression for C at n=2 is given by,

C=142π(Z a o )3/2

Calculation:

The probability is calculated as,

Solving further as,

P=0.08πaor2(2− Zr a o )2e−Zr/ a o|1 4 2π ( Z a o ) 3/2 |2=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3

For r=2ao and Z=1 the above expression is,

P=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3=0.0025ao(2 a o)2(2− 2 a o a o )2e−2 a o/ a o( 1 a o )3=0

Conclusion:

Therefore, the probability of finding the electron in the region between r and r+Δr is 0 .

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Answer 3

Question

Chapter 36, Problem 41P

(a)

To determine

The probability of finding the electron in the region between r and r+Δr .

(a)

Expert Solution

Answer to Problem 41P

The probability of finding the electron in the region between r and r+Δr is 9.20×10−4 .

Explanation of Solution

Given:

The value of Δr is 0.02ao .

The value of r is ao .

The value of n is 2 .

The value of l is 0 .

The value of ml is 0 .

Formula used:

The expression for probabilityis given by,

P=P(r)Δr

The expression for P(r) is given by,

P(r)=4πr2|ψ|2

The expression for ψ at n=2 is given by,

ψ=C(2−Zrao)e−Zr/2ao

The expression for C at n=2 is given by,

C=142π(Z a o )3/2

Calculation:

The probability is calculated as,

Solving further as,

P=0.08πaor2(2− Zr a o )2e−Zr/ a o|1 4 2π ( Z a o ) 3/2 |2=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3

For r=ao and Z=1 the above expression is,

P=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3=0.0025ao( a o)2(2− a o a o )2e− a o/ a o( 1 a o )3=9.2×10−4

Conclusion:

Therefore, the probability of finding the electron in the region between r and r+Δr is 9.20×10−4 .

(b)

To determine

The probability of finding the electron in the region between r and r+Δr .

(b)

Expert Solution

Answer to Problem 41P

The probability of finding the electron in the region between r and r+Δr is 0 .

Explanation of Solution

Given:

The value of Δr is 0.02ao .

The value of r is 2ao .

The value of n is 2 .

The value of l is 0 .

The value of ml is 0 .

Formula used:

The expression for probability is given by,

P=P(r)Δr

The expression for P(r) is given by,

P(r)=4πr2|ψ|2

The expression for ψ at n=2 is given by,

ψ=C(2−Zrao)e−Zr/2ao

The expression for C at n=2 is given by,

C=142π(Z a o )3/2

Calculation:

The probability is calculated as,

Solving further as,

P=0.08πaor2(2− Zr a o )2e−Zr/ a o|1 4 2π ( Z a o ) 3/2 |2=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3

For r=2ao and Z=1 the above expression is,

P=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3=0.0025ao(2 a o)2(2− 2 a o a o )2e−2 a o/ a o( 1 a o )3=0

Conclusion:

Therefore, the probability of finding the electron in the region between r and r+Δr is 0 .

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Answer 4

Question

Chapter 36, Problem 41P

(a)

To determine

The probability of finding the electron in the region between r and r+Δr .

(a)

Expert Solution

Answer to Problem 41P

The probability of finding the electron in the region between r and r+Δr is 9.20×10−4 .

Explanation of Solution

Given:

The value of Δr is 0.02ao .

The value of r is ao .

The value of n is 2 .

The value of l is 0 .

The value of ml is 0 .

Formula used:

The expression for probabilityis given by,

P=P(r)Δr

The expression for P(r) is given by,

P(r)=4πr2|ψ|2

The expression for ψ at n=2 is given by,

ψ=C(2−Zrao)e−Zr/2ao

The expression for C at n=2 is given by,

C=142π(Z a o )3/2

Calculation:

The probability is calculated as,

Solving further as,

P=0.08πaor2(2− Zr a o )2e−Zr/ a o|1 4 2π ( Z a o ) 3/2 |2=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3

For r=ao and Z=1 the above expression is,

P=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3=0.0025ao( a o)2(2− a o a o )2e− a o/ a o( 1 a o )3=9.2×10−4

Conclusion:

Therefore, the probability of finding the electron in the region between r and r+Δr is 9.20×10−4 .

(b)

To determine

The probability of finding the electron in the region between r and r+Δr .

(b)

Expert Solution

Answer to Problem 41P

The probability of finding the electron in the region between r and r+Δr is 0 .

Explanation of Solution

Given:

The value of Δr is 0.02ao .

The value of r is 2ao .

The value of n is 2 .

The value of l is 0 .

The value of ml is 0 .

Formula used:

The expression for probability is given by,

P=P(r)Δr

The expression for P(r) is given by,

P(r)=4πr2|ψ|2

The expression for ψ at n=2 is given by,

ψ=C(2−Zrao)e−Zr/2ao

The expression for C at n=2 is given by,

C=142π(Z a o )3/2

Calculation:

The probability is calculated as,

Solving further as,

P=0.08πaor2(2− Zr a o )2e−Zr/ a o|1 4 2π ( Z a o ) 3/2 |2=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3

For r=2ao and Z=1 the above expression is,

P=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3=0.0025ao(2 a o)2(2− 2 a o a o )2e−2 a o/ a o( 1 a o )3=0

Conclusion:

Therefore, the probability of finding the electron in the region between r and r+Δr is 0 .

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Answer 5

Chapter 36, Problem 41P

(a)

To determine

The probability of finding the electron in the region between r and r+Δr .

(a)

Expert Solution

Answer to Problem 41P

The probability of finding the electron in the region between r and r+Δr is 9.20×10−4 .

Explanation of Solution

Given:

The value of Δr is 0.02ao .

The value of r is ao .

The value of n is 2 .

The value of l is 0 .

The value of ml is 0 .

Formula used:

The expression for probabilityis given by,

P=P(r)Δr

The expression for P(r) is given by,

P(r)=4πr2|ψ|2

The expression for ψ at n=2 is given by,

ψ=C(2−Zrao)e−Zr/2ao

The expression for C at n=2 is given by,

C=142π(Z a o )3/2

Calculation:

The probability is calculated as,

Solving further as,

P=0.08πaor2(2− Zr a o )2e−Zr/ a o|1 4 2π ( Z a o ) 3/2 |2=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3

For r=ao and Z=1 the above expression is,

P=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3=0.0025ao( a o)2(2− a o a o )2e− a o/ a o( 1 a o )3=9.2×10−4

Conclusion:

Therefore, the probability of finding the electron in the region between r and r+Δr is 9.20×10−4 .

(b)

To determine

The probability of finding the electron in the region between r and r+Δr .

(b)

Expert Solution

Answer to Problem 41P

The probability of finding the electron in the region between r and r+Δr is 0 .

Explanation of Solution

Given:

The value of Δr is 0.02ao .

The value of r is 2ao .

The value of n is 2 .

The value of l is 0 .

The value of ml is 0 .

Formula used:

The expression for probability is given by,

P=P(r)Δr

The expression for P(r) is given by,

P(r)=4πr2|ψ|2

The expression for ψ at n=2 is given by,

ψ=C(2−Zrao)e−Zr/2ao

The expression for C at n=2 is given by,

C=142π(Z a o )3/2

Calculation:

The probability is calculated as,

Solving further as,

P=0.08πaor2(2− Zr a o )2e−Zr/ a o|1 4 2π ( Z a o ) 3/2 |2=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3

For r=2ao and Z=1 the above expression is,

P=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3=0.0025ao(2 a o)2(2− 2 a o a o )2e−2 a o/ a o( 1 a o )3=0

Conclusion:

Therefore, the probability of finding the electron in the region between r and r+Δr is 0 .

Answer 6

Chapter 36, Problem 41P

(a)

To determine

The probability of finding the electron in the region between r and r+Δr .

(a)

Expert Solution

Answer to Problem 41P

The probability of finding the electron in the region between r and r+Δr is 9.20×10−4 .

Explanation of Solution

Given:

The value of Δr is 0.02ao .

The value of r is ao .

The value of n is 2 .

The value of l is 0 .

The value of ml is 0 .

Formula used:

The expression for probabilityis given by,

P=P(r)Δr

The expression for P(r) is given by,

P(r)=4πr2|ψ|2

The expression for ψ at n=2 is given by,

ψ=C(2−Zrao)e−Zr/2ao

The expression for C at n=2 is given by,

C=142π(Z a o )3/2

Calculation:

The probability is calculated as,

Solving further as,

P=0.08πaor2(2− Zr a o )2e−Zr/ a o|1 4 2π ( Z a o ) 3/2 |2=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3

For r=ao and Z=1 the above expression is,

P=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3=0.0025ao( a o)2(2− a o a o )2e− a o/ a o( 1 a o )3=9.2×10−4

Conclusion:

Therefore, the probability of finding the electron in the region between r and r+Δr is 9.20×10−4 .

Answer 7

Chapter 36, Problem 41P

(a)

To determine

The probability of finding the electron in the region between r and r+Δr .

Answer 8

(a)

Expert Solution

Answer to Problem 41P

The probability of finding the electron in the region between r and r+Δr is 9.20×10−4 .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The value of Δr is 0.02ao .

The value of r is ao .

The value of n is 2 .

The value of l is 0 .

The value of ml is 0 .

Formula used:

The expression for probabilityis given by,

P=P(r)Δr

The expression for P(r) is given by,

P(r)=4πr2|ψ|2

The expression for ψ at n=2 is given by,

ψ=C(2−Zrao)e−Zr/2ao

The expression for C at n=2 is given by,

C=142π(Z a o )3/2

Calculation:

The probability is calculated as,

Solving further as,

P=0.08πaor2(2− Zr a o )2e−Zr/ a o|1 4 2π ( Z a o ) 3/2 |2=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3

For r=ao and Z=1 the above expression is,

P=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3=0.0025ao( a o)2(2− a o a o )2e− a o/ a o( 1 a o )3=9.2×10−4

Conclusion:

Therefore, the probability of finding the electron in the region between r and r+Δr is 9.20×10−4 .

Answer 13

(b)

To determine

The probability of finding the electron in the region between r and r+Δr .

(b)

Expert Solution

Answer to Problem 41P

The probability of finding the electron in the region between r and r+Δr is 0 .

Explanation of Solution

Given:

The value of Δr is 0.02ao .

The value of r is 2ao .

The value of n is 2 .

The value of l is 0 .

The value of ml is 0 .

Formula used:

The expression for probability is given by,

P=P(r)Δr

The expression for P(r) is given by,

P(r)=4πr2|ψ|2

The expression for ψ at n=2 is given by,

ψ=C(2−Zrao)e−Zr/2ao

The expression for C at n=2 is given by,

C=142π(Z a o )3/2

Calculation:

The probability is calculated as,

Solving further as,

P=0.08πaor2(2− Zr a o )2e−Zr/ a o|1 4 2π ( Z a o ) 3/2 |2=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3

For r=2ao and Z=1 the above expression is,

P=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3=0.0025ao(2 a o)2(2− 2 a o a o )2e−2 a o/ a o( 1 a o )3=0

Conclusion:

Therefore, the probability of finding the electron in the region between r and r+Δr is 0 .

Answer 14

(b)

To determine

The probability of finding the electron in the region between r and r+Δr .

Answer 15

(b)

Expert Solution

Answer to Problem 41P

The probability of finding the electron in the region between r and r+Δr is 0 .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

The value of Δr is 0.02ao .

The value of r is 2ao .

The value of n is 2 .

The value of l is 0 .

The value of ml is 0 .

Formula used:

The expression for probability is given by,

P=P(r)Δr

The expression for P(r) is given by,

P(r)=4πr2|ψ|2

The expression for ψ at n=2 is given by,

ψ=C(2−Zrao)e−Zr/2ao

The expression for C at n=2 is given by,

C=142π(Z a o )3/2

Calculation:

The probability is calculated as,

Solving further as,

P=0.08πaor2(2− Zr a o )2e−Zr/ a o|1 4 2π ( Z a o ) 3/2 |2=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3

For r=2ao and Z=1 the above expression is,

P=0.0025aor2(2− Zr a o )2e−Zr/ a o( Z a o )3=0.0025ao(2 a o)2(2− 2 a o a o )2e−2 a o/ a o( 1 a o )3=0

Conclusion:

Therefore, the probability of finding the electron in the region between r and r+Δr is 0 .

Answer 20

Ch. 36 - Prob. 1P Ch. 36 - Prob. 2P Ch. 36 - Prob. 3P Ch. 36 - Prob. 4P Ch. 36 - Prob. 5P Ch. 36 - Prob. 6P Ch. 36 - Prob. 7P Ch. 36 - Prob. 8P Ch. 36 - Prob. 9P Ch. 36 - Prob. 10P

The probability of finding the electron in the region between r and r + Δ r .

The probability of finding the electron in the region between r and r+Δr .

Answer to Problem 41P

Explanation of Solution

The probability of finding the electron in the region between r and r+Δr .

Answer to Problem 41P

Explanation of Solution

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