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

(a) $2.3 \times 10^{-138}$ is the minimum radius the earth can have for its orbit

(b) $n = 2.5 \times 10^{74}$ is the value of the principal quantum number for the present radius of the earth

Explanation:

It is given that,

Earth’s mass, $m_e$ = 6.0 × 1024 kg

Sun’s mass, $m_s$ = 2.0 × 1030 kg

Distance between the sun and the earth, d = 1.5 × 1111 m

Bohr’s quantization rule says,

Angular momentum, $L = \frac{nh}{2\pi}$

$mvr = \frac{nh}{2\pi}$ ----- (1)

where,

v is the electron velocity  

h = Constant of Planck  

m = Electron mass  

n = Quantum number  

r = Circular orbit’s radius

When both the sides are squared, we obtain

$n^{2}h^2}4\pi^{2} = m_e^{2}v^{2}r^{2}$            ….2

Gravitational attraction force between the sun and the earth acts as the centripetal force.

$F = \frac{m_ev^{2}}{r} = \frac{Gm_em_s}{r_2}$

 $v^{2} = Gm_sr$                ….(3)

When (3) is divided by (2), we obtain

$m_e^{2} r = n^{2}h^{2}4\pi^{2}Gm_s$

(a) The minimum radius the earth can have for its orbit  

For n = 1,

$R = \sqrt{ \frac{ h^{2}}{4\pi^{2} Gm_s m_e^{2}}$

Substitute the values of h, G,$m_s$ and $m_e$

$r=\sqrt{\frac{\left(6.63 \times 10^{-34}\right)^{2}}{4 \times(3.14)^{2} \times\left(6.67 \times 10^{-11}\right) \times\left(6 \times 10^{24}\right)^{2} \times\left(2 \times 10^{30}\right)}}$

$R = 2.3\times 10^{-138} m$

The minimum radius the earth can have for its orbit is $2.3 \times 10^{-138}$ m

(b)  Value of the principal Quantum number n for the present radius

From the equation (2), the principal quantum number (n)’s value is shown as

$n^{2} = m_e^{2} \times r \times 4 \times \pi \times G \times m_s \times h^{2}$

$n = \sqrt{m_e^{2} \times r \times 4 \times \pi \times G \times m_s \times h^{2}}$

Substitute the values of $m_e$ , $m_s$, G, h, r and π

$n=\sqrt{\frac{\left(6 \times 10^{24}\right)^{2} \times\left(1.5 \times 10^{11}\right) \times 4 \times(3.14)^{2} \times\left(6.67 \times 10^{-11}\right) \times\left(2 \times 10^{30}\right)}{6.6 \times 10^{-34}}}$

$n = 2.5 \times 10^{74}$

The value of the principal quantum number is $n = 2.5 \times 10^{74}$ for the present radius

Answer 2

Answer:

Above answer is correct .....