Physics for Scientists and Engineers, Vol. 1
Physics for Scientists and Engineers, Vol. 1
6th Edition
ISBN: 9781429201322
Author: Paul A. Tipler, Gene Mosca
Publisher: Macmillan Higher Education
Question
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Chapter 36, Problem 41P

(a)

To determine

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

(a)

Expert Solution
Check Mark

Answer to Problem 41P

The probability of finding the electron in the region between r and r+Δr is 9.20×104 .

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(2Zrao)eZr/2ao

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

  C=142π(Z a o )3/2

Calculation:

The probability is calculated as,

  P=P(r)Δr=4πr2|ψ|2Δr=4πr2|C( 2 Zr a o )e Zr/ 2 a o |2×0.02ao=0.08πaor2(2 Zr a o )2eZr/ a o|1 4 2π ( Z a o ) 3/2 |2

Solving further as,

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

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

  P=0.0025aor2(2 Zr a o )2eZr/ 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=9.2×104

Conclusion:

Therefore, the probability of finding the electron in the region between r and r+Δr is 9.20×104 .

(b)

To determine

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

(b)

Expert Solution
Check Mark

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(2Zrao)eZr/2ao

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

  C=142π(Z a o )3/2

Calculation:

The probability is calculated as,

  P=P(r)Δr=4πr2|ψ|2Δr=4πr2|C( 2 Zr a o )e Zr/ 2 a o |2×0.02ao=0.08πaor2(2 Zr a o )2eZr/ a o|1 4 2π ( Z a o ) 3/2 |2

Solving further as,

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

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

  P=0.0025aor2(2 Zr a o )2eZr/ a o( Z a o )3=0.0025ao(2 a o)2(2 2 a o a o )2e2 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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