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Chapter 6: Q12E (page 1402)

A hydrogen atom is in a state that has Lz=2hIn the semi classical vector model, the angular momentum vector L鈫�for this state makes an angle胃L=63.4掳with the +z-axis. (a) What is the l quantum number for this state? (b) What is the smallest possible n quantum number for this state?

Short Answer

Expert verified
  1. The value of the orbital quantum number is l=4.
  2. The smallest possible value of n is 5

Step by step solution

01

Important Concepts

Orbital angular momentum in a hydrogen atom is given by

L=l+1h鈥︹赌︹赌︹赌︹赌︹赌�(1)

Where l is the azimuthal quantum number.

The angle the vector L makes with the z-axis is given by

胃=cos-1LzL鈥︹赌︹赌︹赌︹赌︹赌�.(2)

02

Application

Solving the equation (2) for we get,

L=Lz肠辞蝉胃

Plug in the given values

L=2hcos63.40L=4.47h

Square both sides of equation (1) and rearranging it we get,

L2=ll+1h2l2+l-L2h2=0l2+l-20=0

Solve for l and we get

l = 4 and l = - 5

Reject the negative value, therefore the orbital quantum number of this state is l=4

Since orbital quantum numbers can have values from 0 to n - 1

So the possible values of n for this state is n鈮�l+1

Hencen鈮�5

Hence the minimum value of n is 5

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