Question

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

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

Given data:

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

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

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

$\mathit{E}\mathbf{\infty }{\mathit{n}}^{\mathbf{2}}$ .....(I)

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

$\mathit{\lambda }\mathbf{=}\frac{\mathbf{h}}{\sqrt{\mathbf{2}\mathbf{m}\mathbf{E}}}$ .....(II)

Here, is the Planck's constant.

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 $E_3$is the highest, followed by $E_2$and finally $E_1$. 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 .