Question

The earth revolves round the sun due to gravitational attraction. Suppose that the sun and the earth are point particles with their existing masses and that Bohr's quantization rule for angular momentum is valid in the case of gravitation. (a) Calculate the minimum radius the earth can have for its orbit. (b) What is the value of the principal quantum number n for the present radius? Mass of the earth = 6.0 × 10⁻²⁴ kg. Mass of the sun = 2.0 × 10³⁰ kg, earth-sun distance = 1.5 × 10¹¹ m.

Answer 1

Given:
Mass of the earth, m_e = 6.0 × 10²⁴ kg
Mass of the sun, m_s = 2.0 × 10³⁰ kg
Distance between the earth and the sun, d = 1.5 × 11¹¹ m

According to the Bohr's quantization rule,

Angular momentum, L = $\frac{nh}{2\mathrm{\pi }}$
$\Rightarrow mvr=\frac{nh}{2\mathrm{\pi }}$ ....(1)
Here,
n = Quantum number
h = Planck's constant
m = Mass of electron
r = Radius of the circular orbit
v = Velocity of the electron

Squaring both the sides, we get
$m_{\mathrm{e}}^2{v}^2{r}^2=\frac{n^2{h}^2}{4{\mathrm{\pi }}^2}....\left(2\right)$
Gravitational force of attraction between the earth and the sun acts as the centripetal force.
$F=\frac{G{m}_{\mathrm{e}}{m}_{\mathrm{s}}}{r^2}=\frac{m_e{v}^2}{r}$
$\Rightarrow v^2=\frac{G{m}_s}{r}$ ....(3)
Dividing (2) by (3), we get
$m_{\mathrm{e}}^{2}r=\frac{n^2{h}^2}{4{\mathrm{\pi }}^{2}G{m}_{\mathrm{s}}}$

(a) For n = 1,

$$\begin{gathered} r=\sqrt{\frac{h^2}{4{\mathrm{\pi }}^{2}G{m}_{\mathrm{s}}{m}_{\mathrm{e}}^2}} \\ r=\sqrt{\frac{{\left(6.63\times {10}^{-34}\right)}^2}{4\times {\left(3.14\right)}^2\times \left(6.67\times {10}^{-11}\right)\times {\left(6\times {10}^{24}\right)}^2\times \left(2\times {10}^{30}\right)}} \\ r=2.29\times {10}^{-138}\mathrm{m} \\ r=2.3\times {10}^{-138}\mathrm{m} \end{gathered}$$

(b)
From (2), the value of the principal quantum number (n) is given by
$$\begin{gathered} n^2=\frac{m_{\mathrm{e}}^2\times r\times 4\times \mathrm{\pi }\times \mathrm{G}\times m_{\mathrm{s}}}{h^2} \\ \Rightarrow n=\sqrt{\frac{m_{\mathrm{e}}^2\times r\times 4\times \mathrm{\pi }\times \mathrm{G}\times m_{\mathrm{s}}}{h^2}} \end{gathered}$$
$$\begin{gathered} n=\sqrt{\frac{{\left(6\times {10}^{24}\right)}^2\times \left(1.5\times {10}^{11}\right)\times 4\times {\left(3.14\right)}^2\times \left(6.67\times {10}^{-11}\right)\times \left(2\times {10}^{30}\right)}{{\left(6.6\times {10}^{-34}\right)}^2}} \\ n=2.5\times {10}^{74} \end{gathered}$$