An electron is in a state with I = 3 . (a) What multiple of gives the magnitude of L? (b) What multiple of μBgives the magnitude of μ? (c) What is the largest possible value of m1, (d) what multiple of ħgives the corresponding value oflocalid="1661917347426" L, and (e) what multiple of μBgives the corresponding value of μorb,z? (f) What is the value of the semi-classical angle θbetween the directions of LzandL? What is the value of angle θfor (g) the second largest possible value of mIand (h) the smallest (that is, most negative) possible value of mI?

Short Answer

Expert verified

a) Theħmultiple of magnitude of Lis3.46.

b) TheμBmultiple of magnitude of μis3.46 .

c) The largest possible value ofmIis3.

d) The corresponding value ofLzis3.

e) The corresponding value ofμorb,zis-3.

f) The value of the semi-classical angle θbetween the directions LzandLis30° .

g) The value of angle θfor the second largest possible value of mIis54.7° .

h) The value of the angle θfor the smallest possible value of m1is150° .

Step by step solution

01

The given data:

An electron is in state with value I = 3.

02

Understanding the concept of orbital angular quantum values:

Orbital angular momentum quantum number is 'I' . The orbital angular momentum quantum number determines the shape of the orbital and hence the angular distribution. The value of I is dependent on the principal quantum number n .

Using the given value of the orbital angular quantum number, get the magnitude of the orbital angular momentum and dipole moment. As the range of m varies from -I to +I, the maximum value is given by the most positive value of I . Similarly using the concept of the orbital momentum and dipole moment projection along the z-axis, define the required values using the given value in the formula. The semi-classical angle between two vectors can be given using the cosine angle between them.

Formulas:

The magnitude of the orbital angular momentum in terms of ħis,

L=II+1ħ ….. (1)

The magnitude of the orbital dipole moment in terms of Bohr’s magnetron μB,

μorb=II+1μB ….. (2)

The z-component of the orbital angular momentum is,

Lz=mlħ ….. (3)

The z-component of the orbital magnetic dipole moment is,

μorb,z=-mIμB ….. (4)

The semi-classical angle between a vector and its z-component is,

θ=cos-1aza ….. (5)

03

(a) Calculation of the magnitude of orbital angular momentum:

Define the value of the magnitude of the orbital angular momentum by substituting 3 for I in equation (1) as below.

L=33+1ħ=12ħ=3.46ħ

Hence, the ħmultiple value of the orbital angular momentum is 3.46.

04

(b) Calculation of the orbital magnetic dipole moment:

Calculate the value of the orbital magnetic dipole moment ny substituting 3 for I into equation (2) as below.

μorb=33+1μB=12μB=3.46μB

Hence, the μBmultiple value of the dipole moment is 3.46.

05

(c) Calculation of the largest value of mI :

Using the concept of range of Lz=mIħfor I values, get the largest value of the mIas follows:

mI=+I=3

Hence, the largest value of mIis 3 .

06

(d) Calculation of the z-component of the orbital angular momentum:

Using the largest value of mI=3in equation (3), obtain the ħmultiple value of z-component of the orbital angular momentum as follows:

Lz=3ħ

Hence, the ħmultiple value of the z-component is 3 .

07

(e) Calculation of the z-component of the orbital magnetic dipole moment:

Using the largest value of mI=3in equation (3), determine the μBmultiple value of z-component of the orbital angular momentum as follows:

μorb.z=-3μB

Hence, the μBmultiple value of the z-component is -3.

08

(f) Calculation of the angle between the orbital angular momentum and its z-component:

Using the value of Land from part (a) and (d) calculations, get the semi-classical angle between the two directions Lzand Lusing equation (5) as follows:

θ=cos-13ħ12ħ=cos-132=30°

Hence, the value of the angle is 30°.

09

(g) Calculation of the angle for the second largest value of m1 :

Using the concept, we can say that the second largest value of m1is+2.

Thus, using this value in equation (3), determine the component value as below.

Lz=2ħ

Now, the angle between the orbital angular momentum and the second largest value of is given using the above value and value of angular momentum from part (a) calculations in equation (5) as follows:

θ=cos-12ħ12ħ=cos-113=54.7°

Hence, the value of the angle is 54.7°.

10

(h) Calculation of the angle for the smallest value of  :

Using the concept, we can say that the smallest value of mIis-3 .

Thus, using this value in equation (3), get the component value as:

Lz=-3ħ

Now, the angle between the orbital angular momentum and the second largest value of is given using the above value and value of angular momentum from part (a) calculations in equation (5) as follows:

θ=cos-1-3ħ12ħ=cos-1-32=150°

Hence, the value of the angle is 150°.

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