Energy level diagram for one-dimensional box.

Question

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Answer 1

Question

Chapter 41, Problem 7P

(a)

To determine

Energy level diagram for one-dimensional box.

(a)

Expert Solution

Answer to Problem 7P

The energy level diagram is given below:

Physics for Scientists and Engineers with Modern Physics, Technology Update, Chapter 41, Problem 7P , additional homework tip 1

Explanation of Solution

The length of the box is 0.100 nm.

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 6.626×10−34 Js for h, 1 for n, 0.100 nm for L and 9.1×10−31 kg for m in (I) to find En for the first level.

En=12(6.626×10−34 kgm2s−1)28(9.1×10−31 kg)(0.100 nm)2=12(6.626×10−34 kgm2s−1)28(9.1×10−31 kg)(0.100×10−9 m)2⋅1 eV1.602×10−19 kgm2s−2=37.7 eV

Substitute 6.626×10−34 Js for h, 2 for n, 0.100 nm for L and 9.1×10−31 kg for m in (I) to find En for the second level.

En=22(6.626×10−34 kgm2s−1)28(9.1×10−31 kg)(0.100 nm)2=22(6.626×10−34 kgm2s−1)28(9.1×10−31 kg)(0.100×10−9 m)2⋅1 eV1.602×10−19 kgm2s−2=151 eV

Substitute 6.626×10−34 Js for h, 3 for n, 0.100 nm for L and 9.1×10−31 kg for m in (I) to find En for the third level

En=32(6.626×10−34 kgm2s−1)28(9.1×10−31 kg)(0.100 nm)2=32(6.626×10−34 kgm2s−1)28(9.1×10−31 kg)(0.100×10−9 m)2⋅1 eV1.602×10−19 kgm2s−2=339 eV

Substitute 6.626×10−34 Js for h, 4 for n, 0.100 nm for L and 9.1×10−31 kg for m in (I) to find En for the fourth level

En=42(6.626×10−34 kgm2s−1)28(9.1×10−31 kg)(0.100 nm)2=42(6.626×10−34 kgm2s−1)28(9.1×10−31 kg)(0.100×10−9 m)2⋅1 eV1.602×10−19 kgm2s−2=603 eV

The energy level diagram is given below:

Physics for Scientists and Engineers with Modern Physics, Technology Update, 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

Answer to Problem 7P

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

Explanation of Solution

The minimum frequency for photoemission corresponds to the cutoff wavelength.

Write the expression for wavelength

λ=hc|E−E′| (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, 4→3,

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

λ=1240 eV⋅nm603 eV−339 eV=4.71 nm

For transition, 4→2,

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

λ=1240 eV⋅nm603 eV−151 eV=2.75 nm

For transition, 4→1,

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

λ=1240 eV⋅nm603 eV−37.7 eV=2.20 nm

For transition, 3→2,

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

λ=1240 eV⋅nm339 eV−151 eV=6.59 nm

For transition, 3→1,

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

λ=1240 eV⋅nm339 eV−37.7 eV=4.12 nm

For transition, 2→1,

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

λ=1240 eV⋅nm151 eV−37.7 eV=11.0 nm

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

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