Chapter 28, Problem 44GP

(a)

To determine

The quantum number $\mathrm{l}$ for the orbital angular momentum of the Earth about the Sun.

Answer to Problem 44GP

  $\mathrm{l}=2.5\times {10}^{74}$

Explanation of Solution

Given info:

The mass of Earth is ${\text{M}}_{\text{earth}}\text{=}\text{5}\text{.98}\text{×}{\text{10}}^{\text{24}}\text{kg}$

Distance between Earth and Sun is $\mathrm{R}=\text{1}{\text{.50×10}}^{\text{11}}\text{m}$ .

Revolution period of Earth, $\mathrm{T}=\text{3}\text{.16}{\text{×10}}^{\text{7}}\text{s}$

Formula used:

The orbital quantum number $\mathrm{l}$ can take on values from $0$ to $\text{n-1}$ .

The actual magnitude of the angular momentum $\text{L}$ is related to the quantum $\mathrm{l}$ by

  $\text{L = }\mathrm{\hslash }{\left[\mathrm{l}\left(\mathrm{l}+1\right)\right]}^{\frac{1}{2}}$

Where, $\mathrm{l}$ is the orbital quantum number and $\mathrm{\hslash }$ is reduced Planck’s constant, $1.055\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}$

Calculation:

Calculate the quantum number for the orbital angular momentum from

  ${\text{L = M}}_{\text{earth}}\text{vR=}\frac{{\text{M}}_{\text{earth}}{\text{2πR}}^{\text{2}}}{\text{T}}\text{=}\mathrm{\hslash }{\left[\mathrm{l}\left(\mathrm{l}\text{ + 1}\right)\right]}^{\frac{\text{1}}{\text{2}}}$

  $\frac{\left(\text{5}\text{.98}\text{×}{\text{10}}^{\text{24}}\text{kg}\right)\text{2π}{\left(\text{1}{\text{.50×10}}^{\text{11}}\text{m}\right)}^{\text{2}}}{\left(\text{3}\text{.16}{\text{×10}}^{\text{7}}\text{s}\right)}\text{=}\left(\text{1}\text{.055}\text{×}{\text{10}}^{\text{-34}}\text{J}\cdot \text{s}\right){\left[\mathrm{l}\left(\mathrm{l}\text{+}\text{1}\right)\right]}^{\frac{\text{1}}{\text{2}}}$

This gives $\mathrm{l}=2.5\times {10}^{74}$

(b)

To determine

The number of possible orientations for the plane of Earth’s orbit.

Answer to Problem 44GP

  $5.0\times {10}^{74}$

Explanation of Solution

Given info: The orbital quantum number, $\mathrm{l}=2.5\times {10}^{74}$

Formula used:

The orbital quantum number $\mathrm{l}$ can take on values from $0$ to $\text{n-1}$ .

If $\mathrm{l}$ is an angular quantum number of sub shell then max electrons it can hold: $\text{N=}2\left(2\mathrm{l}+1\right)$

Where, $\mathrm{l}$ is the orbital quantum number.

Calculation:

The value of ${\mathrm{m}}_{\mathrm{l}}$ can range from $-\mathrm{l}$ to $+\mathrm{l}$ or $2\mathrm{l}+1$ values.

So, the number of orientations is $\text{N=2}\mathrm{l}\text{+}\text{1}\text{=}\text{2}\left(\text{2}\text{.5}\text{×}{\text{10}}^{\text{74}}\right)\text{+1=5}{\text{.0×10}}^{\text{74}}$