Chapter 9, Problem 47E

(a)

To determine

The de Broglie wavelength of the earth.

Answer to Problem 47E

The de Broglie wavelength of the earth is $\text{3}\text{.68}\times {\text{10}}^{-63}\text{m}$.

Explanation of Solution

Given Info: The mass of earth is $6\times {10}^{24}\text{kg}$ , orbital speed is $3\times {10}^4\text{m}/\text{s}$ and value of h is $6.626\times {10}^{-34}\text{J}\cdot \text{s}$.

Write the formula to calculate the de Broglie wavelength of the earth.

$\lambda =\frac{h}{mv}$

Here,

$\lambda$ is the de Broglie wavelength of the earth

m is the mass of the earth

v is the orbital velocity

Substitute $6\times {10}^{24}\text{kg}$ for m, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for h and $3\times {10}^4\text{m}/\text{s}$ for v in the above equation to calculate $\lambda$.

$\begin{array}{c}\lambda =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}}{\left(6\times {10}^{24}\text{kg}\right)\left(3\times {10}^4\text{m}/\text{s}\right)}\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}}{18\times {10}^{28}\text{kg}\cdot \text{m}/\text{s}}\\ =0.368\times {10}^{-62}\text{m}\\ =\text{3}\text{.68}\times {\text{10}}^{-63}\text{m}\end{array}$

Conclusion:

Therefore, the de Broglie wavelength of the earth is $\text{3}\text{.68}\times {\text{10}}^{-63}\text{m}$.

(b)

To determine

The quantum number of the earth’s orbit.

Answer to Problem 47E

The quantum number of the earth’s orbit is $2.55\times {10}^{74}$.

Explanation of Solution

Given Info: The circumference of the earth is $9.4\times {10}^{11}\text{m}$ and de Broglie wavelength is $\text{3}\text{.68}\times {\text{10}}^{-63}\text{m}$.

Write the expression for the relation between the circumference and wavelength.

$n\lambda =C$

Here,

n is the quantum number of earth’s orbit

C is the circumference of the earth’s orbit

Substitute $9.4\times {10}^{11}\text{m}$ for C and $\text{3}\text{.68}\times {\text{10}}^{-63}\text{m}$ for $\lambda$ in the above equation to calculate n.

$\begin{array}{c}n\left(\text{3}\text{.68}\times {\text{10}}^{-63}\text{m}\right)=9.4\times {10}^{11}\text{m}\\ \text{n}=\frac{9.4\times {10}^{11}\text{m}}{\text{3}\text{.68}\times {\text{10}}^{-63}\text{m}}\\ =2.55\times {10}^{74}\end{array}$

Conclusion:

Therefore, the quantum number of the earth’s orbit is $2.55\times {10}^{74}$.

(c)

To determine

Does the quantum consideration has any major role on the earth’s orbital motion.

Answer to Problem 47E

The mass of the earth is too big and hence the de Broglie wavelength is too small to detect therefore quantum mechanics has no role in the case of earth’s orbit motion.

Explanation of Solution

Write the formula to calculate the de Broglie wavelength of the earth.

$\lambda =\frac{h}{mv}$

Since the de Broglie wavelength of any object is inversely proportional to the mass of the object. As mass increases the observed de Broglie wavelength is too small to detect. Here the mass of the earth is too big and hence too small the de Broglie wavelength.

In addition to that the value of plank’s constant h has the order of ${10}^{-34}$ would again decrease the magnitude of the de Broglie wavelength. Thus, the quantum mechanics has no role on the earth’s orbital motion.

Conclusion:

Therefore, the mass of the earth is too big and hence the de Broglie wavelength is too small to detect therefore quantum mechanics has no role in the case of earth’s orbit motion.