PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
6th Edition
ISBN: 9781429206099
Author: Tipler
Publisher: MAC HIGHER
Question
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Chapter 40, Problem 67P

(a)

To determine

The wavelength of a particle in the ground state of a one-dimensional infinite square well.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The length of the one-dimensional infinite square well is 2.00fm .

Formula used:

Write the expression for the wavelength of a particle in the ground state of a one-dimensional infinite square well.

  λ=2L   ........ (1)

Here, λ is the wavelength of the particle and L is the length of the one-dimensional infinite square well.

Calculation:

Substitute 2.00fm for L in equation (1).

  λ=2(2.00fm)=4.00fm

Conclusion:

Thus, the wavelength of a particle in the ground state of a one-dimensional infinite square well is 4.00fm .

(b)

To determine

The momentum of a particle in the ground state of a one-dimensional infinite square well.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The length of the one-dimensional infinite square well is 2.00fm .

Formula used:

Write the expression for the wavelength of a particle in the ground state of a one-dimensional infinite square well.

  λ=2L   ........ (1)

Here, λ is the wavelength of the particle and L is the length of the one-dimensional infinite square well.

Write the de Broglie relation for the momentum.

  p=hλ

Here, p is the momentum of the particle and h is the Planck’s constant.

Multiplying numerator and denominator by c in the above expression.

  p=hcλc   ........ (2)

Here, c is the speed of light.

Calculation:

Substitute 2.00fm for L in equation (1).

  λ=2(2.00fm)=4.00fm

Substitute 4.00fm for λ and 1240MeVfm for hc in equation (2).

  p=( 1240MeVfm)( 4.00fm)c=310MeV/c

Conclusion:

Thus, the momentum of a particle in the ground state of a one-dimensional infinite square well is 310MeV/c .

(c)

To determine

To show: The total energy of an electron in the ground state of a one-dimensional infinite square well is approximately pc .

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

The length of the one-dimensional infinite square well is 2.00fm .

Formula used:

Write the expression for the wavelength of a particle in the ground state of a one-dimensional infinite square well.

  λ=2L   ........ (1)

Here, λ is the wavelength of the particle and L is the length of the one-dimensional infinite square well.

Write the de Broglie relation for the momentum.

  p=hλ

Here, p is the momentum of the particle and h is the Planck’s constant.

Multiplying numerator and denominator by c in the above expression.

  p=hcλc   ........ (2)

Here, c is the speed of light.

Write the expression for the total energy of the electron.

  E2=E02+p2c2=p2c2(1+ E 0 2 p 2 c 2 )   ........ (3)

Here, E is the total energy of the electron and E0 is the rest energy of the electron.

Calculation:

Substitute 2.00fm for L in equation (1).

  λ=2(2.00fm)=4.00fm

Substitute 4.00fm for λ and 1240MeVfm for hc in equation (2).

  p=( 1240MeVfm)( 4.00fm)c=310MeV/c

Substitute 310MeV/c for p in (1+E02p2c2) .

  (1+ E 0 2 p 2 c 2 )=(1+ E 0 2 ( 310MeV/c ) 2 c 2 )E02<<p2c2and(1+ E 0 2 p 2 c 2 )1

Substitute 1 for (1+E02p2c2) in equation (3).

  E2p2c2

Simplify the above expression for the total energy of the electron.

  Epc

Conclusion:

Thus, the total energy of an electron in the ground state of a one-dimensional infinite square well is approximately pc .

(d)

To determine

The kinetic energy of an electron in the ground state of a one-dimensional infinite square well.

(d)

Expert Solution
Check Mark

Explanation of Solution

Given:

The length of the one-dimensional infinite square well is 2.00fm .

Formula used:

Write the expression for the wavelength of a particle in the ground state of a one-dimensional infinite square well.

  λ=2L   ........ (1)

Here, λ is the wavelength of the particle and L is the length of the one-dimensional infinite square well.

Write the de Broglie relation for the momentum.

  p=hλ

Here, p is the momentum of the particle and h is the Planck’s constant.

Multiplying numerator and denominator by c in the above expression.

  p=hcλc   ........ (2)

Here, c is the speed of light.

Write the expression for the total energy of the electron.

  E2=E02+p2c2=p2c2(1+ E 0 2 p 2 c 2 )   ........ (3)

Here, E is the total energy of the electron and E0 is the rest energy of the electron.

Calculation:

Substitute 2.00fm for L in equation (1).

  λ=2(2.00fm)=4.00fm

Substitute 4.00fm for λ and 1240MeVfm for hc in equation (2).

  p=( 1240MeVfm)( 4.00fm)c=310MeVc

Substitute 310MeV/c for p in (1+E02p2c2) .

  (1+ E 0 2 p 2 c 2 )=(1+ E 0 2 ( 310 MeV c ) 2 c 2 )E02<<p2c2and(1+ E 0 2 p 2 c 2 )1

Substitute 1 for (1+E02p2c2) in equation (3).

  E2p2c2

Simplify the above expression for the total energy of the electron.

  Epc

Write the expression for the kinetic energy of an electron in the ground state of a one-dimensional infinite square well.

  K=EE0E

Here, K is the kinetic energy of an electron in the ground state of the well.

Substitute pc for E in the above expression.

  K=pc   ........ (4)

Substitute 310MeV/c for p in equation (4).

  K=(310 MeVc)c=310MeV

Conclusion:

Thus, the kinetic energy of an electron in the ground state of a one-dimensional infinite square well is 310MeV .

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