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From a visual inspection of Fig. 39-8, rank the quantum numbers of the three quantum states according to the de Broglie wavelength of the electron, greatest first.

Short Answer

Expert verified

According to de Broglie wavelength, the electron rank of the quantum states with gratest first is n = 1, n = 2, n = 3.

Step by step solution

01

Given data:

Three wave functions corresponding to quantum states with quantum numbers n = 1,2,3

02

Energy of electrons in potential well and de-Broglie wavelength:

The energy of an electron in a potential well varies with quantum number as

E∞n2 .....(I)

The de-Broglie wavelength of an electron of mass having energy is

λ=h2mE .....(II)

Here, is the Planck's constant.

03

Determining the order of the de Broglie wavelength of the electrons:

From equation (I) the energy of the electron in the n = 3 state, that is E3is the highest, followed by E2and finally E1. From equation (II), the de-Broglie wavelength is inversely proportional to energy.

Thus, the particle with the lowest energy will have the highest wavelength. Hence the electron at n = 1 has the highest wavelength, followed by the one at n = 2 and the one at n = 3 .

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Most popular questions from this chapter

Figure 39-29 a shows a thin tube in which a finite potential trap has been set up where V2=0V. An electron is shown travelling rightward toward the trap, in a region with a voltage of V1=-9.00V, where it has a kinetic energy of 2.00 eV. When the electron enters the trap region, it can become trapped if it gets rid of enough energy by emitting a photon. The energy levels of the electron within the trap are E1=1.0,E2=2.0, and E3=4.0eV, and the non quantized region begins at E4=-9.0eVas shown in the energylevel diagram of Fig. 39-29b. What is the smallest energy such a photon can have?

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