Modern Physics, 3rd Edition
Modern Physics, 3rd Edition
3rd Edition
ISBN: 9780534493394
Author: Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher: Cengage Learning
Question
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Chapter 6, Problem 17P

(a)

To determine

The sketch of wave functions and probability density for n=1 and n=2 states.

(a)

Expert Solution
Check Mark

Answer to Problem 17P

The sketch of wave functions and probability density for n=1 and n=2 states is plotted below.

Explanation of Solution

The sketch of wave function is plotted below.

Modern Physics, 3rd Edition, Chapter 6, Problem 17P , additional homework tip  1

The sketch of probability density is plotted below.

Modern Physics, 3rd Edition, Chapter 6, Problem 17P , additional homework tip  2

Conclusion:

The sketch of wave functions and probability density for n=1 and n=2 states is plotted above.

(b)

To determine

The probability of finding the electron between 0.15 nm and 0.35 nm for n=1 state.

(b)

Expert Solution
Check Mark

Answer to Problem 17P

The probability of finding the electron for n=1 state is 0.200 .

Explanation of Solution

Write the expression for wave function.

  ψ(x)=2Lsin(nπxL)        (I)

Here, A is normalization constant, x is position of box, n is energy state, L is the length of box and ψ(x) is the wave function.

Write the expression for probability.

  P=0L/3|ψ|2dx        (II)

Conclusion:

Substitute 2Lsin(nπxL) for ψ(x)  and substitute 1 for n in equation (II).

  P=(2L)LiLfsin2(πxL)dx

Substitute 1.5Ao for Li , 3.5Ao for Lf and 10Ao for L in above equation and simplify.

  P=110[x5πsin(πx5)]1.53.5=15{3.55πsin[π(3.5)5]1.5+5πsin[π(1.5)5]}=0.200

Thus, the probability of finding the electron for n=1 state is 0.200 .

(c)

To determine

The probability of finding the electron between 0.15 nm and 0.35 nm for n=1 state.

(c)

Expert Solution
Check Mark

Answer to Problem 17P

The probability of finding the electron for n=2 state is 0.351 .

Explanation of Solution

Conclusion:

Substitute 2Lsin(nπxL) for ψ(x)  and substitute 2 for n in equation (II).

  P=(2L)LiLfsin2(2πxL)dx

Substitute 1.5Ao for Li , 3.5Ao for Lf and 10Ao for L in above equation and simplify.

  P=110[x52πsin(0.4πx)]1.53.5=15{3.552πsin(1.4π)1.5+52πsin[0.6π]}=0.351

Thus, the probability of finding the electron for n=2 state is 0.351 .

(d)

To determine

The energies in electron volts for n=1 and n=2 states.

(d)

Expert Solution
Check Mark

Answer to Problem 17P

The energy for n=1 state is 37.7eV and the energy for n=2 state is 151eV .

Explanation of Solution

Write the expression for energy of particle.

  E=n2h28mL2        (III)

Here, n is the energy level, h is the Planck’s constant, m is the mass of particle, L is the length of box and E is the energy of particle.

Conclusion:

Substitute 6.63×1034Js for h , 1010m for L, 9.11×1031kg for m and 1 for n in equation (III).

  E1=(1)2(6.63×1034Js)28(9.11×1031kg)(1010m)2=37.7eV

Substitute 6.63×1034Js for h , 1010m for L, 9.11×1031kg for m and 2 for n in equation (III).

  E2=(2)2(6.63×1034Js)28(9.11×1031kg)(1010m)2=151eV

Thus, the energy for n=1 state is 37.7eV and the energy for n=2 state is 151eV .

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