Chapter 42, Problem 76AP

(a)

To determine

Prove that the quantized radii of Earth’s orbit is $r=\frac{n^2{\hslash }^2}{G{M}_s{M}_E^2}$.

Answer to Problem 76AP

It is proved that the quantized radii of Earth’s orbit is $r=\frac{n^2{\hslash }^2}{G{M}_s{M}_E^2}$.

Explanation of Solution

The gravitational force between Earth and sun is balanced by the centripetal force of rotational motion.

Write the relation between gravitational force between Earth and sun and the centripetal force at equilibrium.

  $G\frac{M_S{M}_E}{r^2}=M_E\frac{v^2}{r}$                                                                                           (I)

Here, $G$ is the gravitational constant, $M_s$ is the mass of sun, $M_E$ is the mass of Earth, $r$ is the distance between Earth and sun, and $v$ is the linear speed of earth.

Write the equation for quantization of angular momentum of Earth.

  $M_{E}vr=n\hslash \left(n=1,2,3,\dots \right)$

Rewrite the above relation in terms of $v$.

    $v=\frac{n\hslash }{M_{E}r}$

Rewrite equation (I) by substituting the above relation for $v$.

  $G\frac{M_S{M}_E}{r^2}=M_E\frac{{\left(\frac{n\hslash }{M_{E}r}\right)}^2}{r}$

Rewrite the above relation inn terms of $r$.

  $r=\frac{n^2{\hslash }^2}{G{M}_S{M}_E^2}$                                                                                     (II)

Therefore, it is proved that the quantized radii of Earth’s orbit are $r=\frac{n^2{\hslash }^2}{G{M}_s{M}_E^2}$.

(b)

To determine

The numerical value of $n$ for Sun-Earth system.

Answer to Problem 76AP

The numerical value of $n$ is $2.53\times {10}^{74}$.

Explanation of Solution

Rewrite equation (II) in terms of $n$.

  $n=\sqrt{G{M}_{S}r}\frac{M_E}{\hslash }$

Conclusion:

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $1.99\times {10}^{30}\text{kg}$ for $M_S$, $1.496\times {10}^{11}\text{m}$ for $r$, $5.98\times {10}^{24}\text{kg}$ for $M_E$, and $1.055\times {10}^{-34}\text{ J}\cdot \text{s}$ for $\hslash$ in the above equation to find $n$.

  $\begin{array}{c}n=\sqrt{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}\right)\left(1.99\times {10}^{30}\text{kg}\right)\left(1.496\times {10}^{11}\text{m}\right)}\frac{\left(5.98\times {10}^{24}\text{kg}\right)}{1.055\times {10}^{-34}\text{ J}\cdot \text{s}}\\ =\left(4.46\times {10}^{15}\text{J}\cdot \text{s}/\text{m}\cdot \text{kg}\right)\left(5.67\times {10}^{58}\text{m}\cdot \text{kg}/\text{J}\cdot \text{s}\right)\\ =2.53\times {10}^{74}\end{array}$

Therefore, the numerical value of $n$ is $2.53\times {10}^{74}$.

(c)

To determine

The distance between orbits corresponding to quantum number $n$ and $n+1$.

Answer to Problem 76AP

The distance between orbits corresponding to quantum number $n$ and $n+1$ is $1.18\times {10}^{-63}\text{ m}$.

Explanation of Solution

Write equation (II) for the radius corresponding to quantum number $n$.

  $r_n=\frac{n^2{\hslash }^2}{G{M}_S{M}_E^2}$                                                                                                    (III)

Here, $r_n$ is the radius of $n^{th}$ orbit.

Rewrite the above equation by substituting $n$ with $n+1$.

  $r_{n+1}=\frac{{\left(n+1\right)}^2{\hslash }^2}{G{M}_S{M}_E^2}$                                                                                                (IV)

Here, $r_{n+1}$ is the radius of ${\left(n+1\right)}^{th}$ orbit.

Subtract equation (III) from equation (IV) to find the separation between orbits.

    $\begin{array}{c}\Delta r=\frac{{\left(n+1\right)}^2{\hslash }^2}{G{M}_S{M}_E^2}-\frac{n^2{\hslash }^2}{G{M}_S{M}_E^2}\\ =\frac{{\hslash }^2}{G{M}_S{M}_E^2}\left(n^2+2n+1-n^2\right)\\ =\frac{\left(2n+1\right){\hslash }^2}{G{M}_S{M}_E^2}\end{array}$

Neglect the value $1$ in the above expression due to the too high value of $n$.

  $\Delta r=\frac{2n{\hslash }^2}{G{M}_S{M}_E^2}$

Conclusion:

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $1.99\times {10}^{30}\text{kg}$ for $M_S$, $1.496\times {10}^{11}\text{m}$ for $r$, $5.98\times {10}^{24}\text{kg}$ for $M_E$, $1.055\times {10}^{-34}\text{ J}\cdot \text{s}$ for $\hslash$ , and $2.53\times {10}^{74}$ for $n$ in the above equation to find $\Delta r$.

  $\begin{array}{c}\Delta r=\frac{2\left(2.53\times {10}^{74}\right){\left(1.055\times {10}^{-34}\text{ J}\cdot \text{s}\right)}^2}{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}\right)\left(1.99\times {10}^{30}\text{kg}\right)\left(5.98\times {10}^{24}\text{kg}\right)}\\ =1.18\times {10}^{-63}\text{ m}\end{array}$

Therefore, the distance between orbits corresponding to quantum number $n$ and $n+1$ is $1.18\times {10}^{-63}\text{ m}$.

(d)

To determine

Importance of results obtained part (b) and (c).

Answer to Problem 76AP

Since the separation between orbits is too small in comparison with the radii of atomic nuclie, it is impossible to observe the quantized orbits of Earth.

Explanation of Solution

The result obtained in part (c) is $1.18\times {10}^{-63}\text{ m}$. The average radius of an atomic nuclei is ${10}^{-15}\text{m}$. It can be seen that the value $1.18\times {10}^{-63}\text{ m}$ is too smaller than ${10}^{-15}\text{m}$, practically it is quite difficult to observe the individual orbits.

Therefore, since the separation between orbits is too small in comparison with the radii of atomic nuclie, it is impossible to observe the quantized orbits of Earth.