A hydrogen atom has l=2. What are the (a) minimum (as a multiple of h ) and (b) maximum values of the quantity Lx2+Ly21/2 ?

Short Answer

Expert verified

Part (a) The minimum value of Lx2+Ly21/2=2

Part (b) The maximum value of Lx2+Ly21/2=6

Step by step solution

01

Step 1. Given information

The equation of total angular momentum,

L2=Lx2+Ly2+Lz2

Here,

L=l(l+1)

Lx2+Ly2=L2-Lz2

Here Lx,Ly,Lz are the components of angular momentum

Lzmax=l

Lzmin=0

02

Step 2. Finding the minimum value of Lx2+Ly212

Lx2+Ly2=L2-Lzmax2

=l(l+1)2-l22

Lx2+Ly21/2=l

Lx2+Ly21/2=2

The minimum value ofLx2+Ly21/2=2

03

Step 3. Finding the maximum value of Lx2+Ly21/2=6ℏ

Lx2+Ly2=L2-Lzmin2

=l(l+1)2-0

Lx2+Ly2=l(l+1)2

Lx2+Ly21/2=l(l+1)

Lx2+Ly21/2=2(2+1)

=6

The maximum value ofLx2+Ly21/2=6

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Most popular questions from this chapter

The 1997Nobel Prize in physics went to Steven Chu, Claude Cohen-Tannoudji, and William Phillips for their development of techniques to slow, stop, and “trap” atoms with laser light. To see how this works, consider a beam of rubidium atoms mass1.4×10-25kg traveling at 500m/safter being evaporated out of an oven. A laser beam with a wavelength of 780nm is directed against the atoms. This is the wavelength of the 5s5ptransition in rubidium, with 5s being the ground state, so the photons in the laser beam are easily absorbed by the atoms. After an average time of 15ns, an excited atom spontaneously emits a 780nmwavelength photon and returns to the ground state.

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