Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Question
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Chapter 41, Problem 7P

(a)

To determine

Energy level diagram for one-dimensional box.

(a)

Expert Solution
Check Mark

Answer to Problem 7P

The energy level diagram is given below:

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University, Chapter 41, Problem 7P , additional homework tip  1

Explanation of Solution

The length of the box is 0.100nm.

Write the expression for energy of one dimensional box

    En=n2h28mL2                                                                                                               (I)

Here, n is the quantum number, h is the Planck’s constant, m  is the mass of electron and L is the length of the box and En is the energy of level n.

Conclusion

Substitute 6.626×1034Js for h, 1 for n, 0.100nm for L and 9.1×1031kg for m in (I) to find En for the first level.

    En=12(6.626×1034kgm2s1)28(9.1×1031kg)(0.100nm)2=12(6.626×1034kgm2s1)28(9.1×1031kg)(0.100×109m)21 eV1.602×1019kgm2s2=37.7eV

Substitute 6.626×1034Js for h, 2 for n, 0.100nm for L and 9.1×1031kg for m in (I) to find En for the second level.

    En=22(6.626×1034kgm2s1)28(9.1×1031kg)(0.100nm)2=22(6.626×1034kgm2s1)28(9.1×1031kg)(0.100×109m)21 eV1.602×1019kgm2s2=151eV

Substitute 6.626×1034Js for h, 3 for n, 0.100nm for L and 9.1×1031kg for m in (I) to find En for the third level

    En=32(6.626×1034kgm2s1)28(9.1×1031kg)(0.100nm)2=32(6.626×1034kgm2s1)28(9.1×1031kg)(0.100×109m)21 eV1.602×1019kgm2s2=339eV

Substitute 6.626×1034Js for h, 4 for n, 0.100nm for L and 9.1×1031kg for m in (I) to find En for the fourth level

    En=42(6.626×1034kgm2s1)28(9.1×1031kg)(0.100nm)2=42(6.626×1034kgm2s1)28(9.1×1031kg)(0.100×109m)21 eV1.602×1019kgm2s2=603eV

The energy level diagram is given below:

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University, Chapter 41, Problem 7P , additional homework tip  2

(b)

To determine

The wavelength of emitted photons during transitions in these four levels.

(b)

Expert Solution
Check Mark

Answer to Problem 7P

The frequencies of the emitted photon are 4.71nm, 2.75nm,2.20nm, 6.59nm, 4.12nm and 11.0nm.

Explanation of Solution

The minimum frequency for photoemission corresponds to the cutoff wavelength.

Write the expression for wavelength

    λ=hc|EE|                                                                                                                (II)

Here, λ is the wavelength, c is the speed of light, E is the higher level energy and E is the lower level energy.

Conclusion

For transition, 43,

Substitute 1240 eVnm for hc, 603eV for E and 339eV for E in (II) to find λ

    λ=1240 eVnm603eV339eV=4.71nm

For transition, 42,

Substitute 1240 eVnm for hc, 603eV for E and 151eV for E in (II) to find λ

    λ=1240 eVnm603eV151eV=2.75nm

For transition, 41,

Substitute 1240 eVnm for hc, 603eV for E and 37.7eV for E in (II) to find λ

    λ=1240 eVnm603eV37.7eV=2.20nm

For transition, 32,

Substitute 1240 eVnm for hc, 339eV for E and 151eV for E in (II) to find λ

    λ=1240 eVnm339eV151eV=6.59nm

For transition, 31,

Substitute 1240 eVnm for hc, 339eV for E and 37.7eV for E in (II) to find λ

    λ=1240 eVnm339eV37.7eV=4.12nm

For transition, 21,

Substitute 1240 eVnm for hc, 151eV for E and 37.7eV for E in (II) to find λ

    λ=1240 eVnm151eV37.7eV=11.0nm

Thus, the frequencies of the emitted photon are 4.71nm, 2.75nm,2.20nm, 6.59nm, 4.12nm and 11.0nm.

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Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

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