PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 28, Problem 29P

(a)

To determine

The ground state energy of the electron

(a)

Expert Solution
Check Mark

Answer to Problem 29P

The ground state energy of the electron is 0.40eV

Explanation of Solution

The energy of the emitted photon during the transition from first excited state to the ground state is 1.2 eV.

Write the expression for energy difference between ground state and first excited state

ΔE=E2E1                                                               (I)

Here, ΔE is the energy difference, E1 is the ground state energy. E2 is the energy of the first excited state.

Write the relation between En and E1

En=n2E1                                                               (II)

Substituting (II) in (I) and substituting 2 for n

ΔE=(221)E1                                                         (III)

Rearranging (III)

E1=ΔE3                                                     (IV)

Conclusion:

Substituting 1.2 eV for ΔE in (IV)  to find E1

E1=1.2 eV3=0.40eV

Thus, the ground state energy of the electron is 0.40eV

(b)

To determine

The possible energies of the photon when a downward transition occur starting from the second excited state.

(b)

Expert Solution
Check Mark

Answer to Problem 29P

The possible energies of the emitted photons are 3.2 eV, 2.0 eV,1.2 eV.

Explanation of Solution

When the electron starts from the first excited state, the possible downward transitions are 31, 32and 21.

The transitions are given in the energy level diagram given below

PHYSICS, Chapter 28, Problem 29P

Write the expression for energy of the emitted photon during transition

Enn=EnEn                                                     (V)

Here, Enn is the energy of the emitted photon, En is the energy of initial state and En energy of final state.

Substituting the corresponding values of n and n for the above given transition.

E32=E3E2                                                  (VI)

E31=E3E1                                                   (VII)

E21=E2E1                                                   (VIII)

Thus, the possible energies of the emitted photons are E31, E32, E21.

Conclusion

Substituting 3 and 2 for n in (II)  to find E3 and E2 respectively

E3=32E1=9E1

E2=22E1=4E1

 Substituting 9E1 for E3, 4E1 for E2 and 0.40eV for E1 in (VI)

E32=(94)0.40 eV=2.0 ev

Substituting 9E1 for E3, and 0.40eV for E1 in (VII)

E31=(91)0.40 eV=3.2 ev

Substituting 4E1 for E2 and 0.40eV for E1 in (VIII)

E21=(41)0.40 eV=1.2 ev

Thus, the possible energies of the emitted photons are 3.2 eV, 2.0 eV,1.2 eV.

(c)

To determine

The length of the one dimensional box.

(c)

Expert Solution
Check Mark

Answer to Problem 29P

The length of the box is 0.97nm.

Explanation of Solution

The length of the box is 2a0. Here, a0 is the Bohr radius

Write the expression for energy of a particle in a 1D box.

E1=h28mL2                                                            (IX)

Here, m is mass of the electron, h is the Planck’s constant and L is the length of the box.

Rearranging (IX)

L=h28mE1                                                            (X)

Conclusion:

Substitute, 0.40eV for E1, 9.109×1031kg for m and 6.626×1034 Js for h in equation (X)  to find L

L=(6.626×1034 Js)28×9.109×1031kg×(0.40eV)=(6.626×1034 Js)28×9.109×1031kg×(0.401.602×1019 J)=0.97 nm

Thus, the length of the box is 0.97nm.

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Chapter 28 Solutions

PHYSICS

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