Question

(a) What is the magnitude of the angular momentum for an l = 1 electron?

(b) Calculate the magnitude of the electron’s spin angular momentum.

(c) What is the ratio of these angular momenta?

Short answer

a)The magnitude of angular momentum for an object is l = 1 electron is \[1.492 \times {10^{ - 34}}\;{\rm{J}} \cdot {\rm{s}}\].

b)The spin angular momentum of an electron has a value of\[9.1327 \times {10^{ - 35}}\;{\rm{J}} \cdot {\rm{s}}\].

c)The angular momentum to spin angular momentum ratio is 1.633.

Step-by-step solution

Definition of angular momentum and magnitude

The magnitude of angular momentum is given as

\[{\bf{L = }}\left( {\sqrt {{\bf{l}}\left( {{\bf{l + 1}}} \right)} } \right)\left( {\frac{{\bf{h}}}{{{\bf{2\pi }}}}} \right)\]

Here,

L = Angular momentum

l = Angular momentum quantum

h = Planck's constant

The magnitude of electron's spin angular momentum is given as

\[{\bf{S = }}\left( {\sqrt {{\bf{s}}\left( {{\bf{s + 1}}} \right)} } \right)\left( {\frac{{\bf{h}}}{{{\bf{2\pi }}}}} \right)\]

Here,

S = Electron's spin angular momentum

s = Spin quantum number

h = Planck's Constant

Calculating the magnitude of the angular momentum

a)

Magnitude of angular momentum is calculated:

\[\begin{array}{c}L = \left( {\sqrt {1\left( {1 + 1} \right)} } \right)\left( {\frac{{6.63 \times {{10}^{ - 3}}}}{{2\pi }}} \right)\;{\rm{J}} \cdot {\rm{s}}\\L = \sqrt 2 \left( {1.055 \times {{10}^{ - 34}}} \right)\;{\rm{J}} \cdot {\rm{s}}\\L = 1.492 \times {10^{ - 34}}\;{\rm{J}} \cdot {\rm{s}}\end{array}\]

Hence, the magnitude of angular momentum for an l = 1 electron is \[1.492 \times {10^{ - 34}}\;\;{\rm{J}} \cdot {\rm{s}}\].

Step 3: Calculate the magnitude

b)

Magnitude ofspin angular momentum is calculated:

\[\begin{array}{c}S = \left( {\sqrt {\frac{1}{2}\left( {\frac{1}{2} + 1} \right)} } \right)\left( {\frac{{6.63 \times {{10}^{ - 34}}}}{{2\pi }}} \right)\;{\rm{J}} \cdot {\rm{s}}\\S = \left( {\sqrt {\frac{3}{4}} } \right)\left( {\frac{{6.63 \times {{10}^{ - 34}}}}{{2\pi }}} \right)\;{\rm{J}} \cdot {\rm{s}}\\S = 9.1327 \times {10^{ - 35}}\;{\rm{J}} \cdot {\rm{s}}\end{array}\]

Thus, the magnitude of the electron's spin angular momentum is \[9.1327 \times {10^{ - 35}}\;{\rm{J}} \cdot {\rm{s}}\].

Finding the ratio of these angular momenta

c)

Determine the ratio of the angular momentum

\[\begin{array}{c}\frac{L}{s} = \frac{{1.492 \times {{10}^{ - 3}}}}{{9.1327 \times {{10}^{ - 13}}}}\\\frac{L}{s} = 1.633\end{array}\]

Thus, the ratio of angular momentum to spin angular momentum is 1.633.