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Chapter 29, Problem 12P

(a)

To determine

To show that the given wave function is normalized.

(a)

Expert Solution
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Answer to Problem 12P

The given wave function is normalized.

Explanation of Solution

Write the condition for the normalization.

  0|ψ1s(r)|2dV=1                                                                                                       (I)

Here, dV is the elemental volume of the sphere.

Write the expression for the elemental volume.

  dV=4πr2dr

Substitute 1πa03era0 for ψ1s(r) and 4πr2dr for dV in the equation (I) to check the wave function is normalized.

  0|1πa03era0|2r2dr=1(4ππa03)0|era0|2r2dr=1(4ππa03)0e2ra0r2dr=1(4a03)(2(2a0)3)=11=1

The integral 0e2ra0r2dr has the value (2(2a0)3) using the gamma function concept.

Thus, the left hand side and right hand side is equal and the condition for the normalization is satisfied.

Conclusion:

Therefore, the given wave function is normalized.

(b)

To determine

The probability of the electron.

(b)

Expert Solution
Check Mark

Answer to Problem 12P

The probability of the electron is 0.498.

Explanation of Solution

Write the expression to calculate the probability.

  P=a023a02|ψ1s(r)|2dV

Substitute 1πa03era0 for ψ1s(r) and 4πr2dr for dV in the above equation to calculate P.

  P=a023a02|1πa03era0|24πr2drP=4ππa03a023a02|era0|2r2drP=4a03a023a02e2ra0r2dr

Refer the table of integrals to solve the above expression.

  P=4a03(a02)[e2ra0(r2+a0r+a022)]a023a02=2a02[e2a0(3a02)((3a02)2+a0(3a02)+a022)e2a0(a02)((a02)2+a0(a02)+a022)]=2a02(e3(17a024)e1(5a024))=2(0.2110.460)=0.498

Conclusion:

Therefore, the probability of the electron is 0.498.

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Chapter 29 Solutions

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term

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