Chapter 28, Problem 50AP

(a)

To determine

The radii of the earth’s orbit of the hydrogen atom from Bohr model.

Answer to Problem 50AP

The radii of the earth’s orbit of the hydrogen atom from Bohr model is $r_n=\frac{n^2{\hslash }^2}{G{M}_s{M}_E^2}$.

Explanation of Solution

Expression the angular momentum associated with the orbital motion of Earth to satisfy Bohr’s model is,

$\begin{array}{l}{M}_{E}vr=n\hslash \\ v_n=\frac{n\hslash }{M_E{r}_n}\end{array}$ (I)

• $v_n$ is the nth level speed,
• $n$ is the nth energy level
• $M_E$ is the Mass of earth
• $r_n$ is the radius of the nth orbit

The gravitational force and the centripetal force compensate each other. So, the expression will be,

$\begin{array}{l}\frac{M_E{v}^2}{r}=\frac{G{M}_s{M}_E}{r^2}\\ v^2=\frac{G{M}_s}{r}\end{array}$ (II)

• $G$ is Gravitational constant,
• $M_s$ is the sun

Calculating the radius of the orbit by using the equation (I) and (II),

$\begin{array}{l}{\left(\frac{n\hslash }{M_E{r}_n}\right)}^2=\frac{G{M}_s}{r_n}\\ r_n=\frac{n^2{\hslash }^2}{G{M}_s{M}_E^2}\end{array}$

Conclusion:

Therefore, the radius of the earth’s orbit of the hydrogen atom from Bohr model is $r_n=\frac{n^2{\hslash }^2}{G{M}_s{M}_E^2}$.

(b)

To determine

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

Answer to Problem 50AP

The numerical value of $n$ for the Sun-Earth system is $2.54\times {10}^{74}$.

Explanation of Solution

Formula to calculate the numerical value of $n$ is,

$r_n=\frac{n^2{\hslash }^2}{G{M}_s{M}_E^2}$

Substitute $5.98\times {10}^{24}\text{kg}$ for $M_E$, $1.99\times {10}^{30}\text{kg}$ for $M_s$, $6.67\times {10}^{-11}{\text{N-m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for $G$, $1.496\times {10}^{11}\text{m}$ for $r$, $1.05\times {10}^{-34}\text{J-s}$ for $\hslash$, to find the value of $n$,

$\begin{array}{l}\left(1.496\times {10}^{11}\text{m}\right)=\frac{n^2{\left(1.05\times {10}^{-34}\text{J-s}\right)}^2}{\left(6.67\times {10}^{-11}{\text{N-m}}^{\text{2}}\right)\left(1.99\times {10}^{30}\text{kg}\right)\left(5.98\times {10}^{24}\text{kg}\right)}\\ n=2.54\times {10}^{74}\end{array}$

Thus, the numerical value of $n$ for the Sun-Earth system is $2.54\times {10}^{74}$.

Conclusion:

Therefore, the numerical value of $n$ for the Sun-Earth system is $2.54\times {10}^{74}$.

(c)

To determine

The distance between the orbits for the quantum number $n$ and the next orbit out from the sun corresponding to the quantum number $\left(n+1\right)$.

Answer to Problem 50AP

The distance between the orbits for the quantum number $n$ and the next orbit out from the sun corresponding to the quantum number $\left(n+1\right)$ is $1.18\times {10}^{-63}\text{m}$.

Explanation of Solution

Formula to calculate the numerical value of $n$ is,

$\begin{array}{c}\left(\Delta r\right)=r_{n+1}-r_n\\ =\frac{\left[{\left(n+1\right)}^2-n^2\right]{\hslash }^2}{G{M}_s{M}_E^2}\\ =\frac{\left(2n+1\right){\hslash }^2}{G{M}_s{M}_E^2}\\ \approx \frac{2n{\hslash }^2}{G{M}_s{M}_E^2}\end{array}$

Substitute $5.98\times {10}^{24}\text{kg}$ for $M_E$, $1.99\times {10}^{30}\text{kg}$ for $M_s$, $2.54\times {10}^{74}$ for $n$, $6.67\times {10}^{-11}{\text{N-m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for $G$, $1.05\times {10}^{-34}\text{J-s}$ for $\hslash$, to find the value of $n$,

$\begin{array}{c}\left(\Delta r\right)=\frac{2\left(2.54\times {10}^{74}\right){\left(1.05\times {10}^{-34}\text{J-s}\right)}^2}{\left(6.67\times {10}^{-11}{\text{N-m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(1.99\times {10}^{30}\text{kg}\right){\left(5.98\times {10}^{24}\text{kg}\right)}^2}\\ =1.18\times {10}^{-63}\text{m}\end{array}$

Thus, the distance between the orbits for the quantum number $n$ and the next orbit out from the sun corresponding to the quantum number $\left(n+1\right)$ is $1.18\times {10}^{-63}\text{m}$.

Conclusion:

Therefore, the distance between the orbits for the quantum number $n$ and the next orbit out from the sun corresponding to the quantum number $\left(n+1\right)$ is $1.18\times {10}^{-63}\text{m}$.

(d)

To determine

The significance of the computed result and the standard result for the distance.

Answer to Problem 50AP

The computed result is much smaller than the standard result.

Explanation of Solution

The computed result is much smaller than the standard result. The standard value of the radius is in the range of $\sim {10}^{15}$. And the computed result is in the rage of $~{10}^{-63}$.

Thus, the computed result is much smaller than the standard result.

Conclusion:

Therefore, the computed result is much smaller than the standard result