Question

In accordance with the bohr's model the quantum number that characterises the earth revolution around the sun in an orbit of radius 1.5 x 10¹¹ m with orbital speed 3 x 10⁴ m/s is [given mass of the earth = 6 x 10²⁴ kg]

• A. 8.57x 10⁶⁴
• B. 5.98 x 10⁶⁴
• C. 2.57 x 10⁷⁴
• D. 2.57x 10³⁸

Answer 1

The Correct Option is C

According to Bohr’s model, angular momentum is quantized and given by: 

\( mvr = \dfrac{nh}{2\pi} \)

Rearranging to find \( n \):

\( n = \dfrac{2\pi mvr}{h} \)

Substitute the known values:

\[ n = \frac{2\pi \times 6.0 \times 10^{24} \times 3 \times 10^4 \times 1.5 \times 10^{11}}{6.62 \times 10^{-34}} \]

Simplifying the numerator:

\[ 2\pi \times 6.0 \times 3 \times 1.5 = 2\pi \times 27 = 169.65 \quad \text{(approx)} \]

So the numerator becomes:

\[ 169.65 \times 10^{24 + 4 + 11} = 169.65 \times 10^{39} \]

Finally:

\[ n = \frac{169.65 \times 10^{39}}{6.62 \times 10^{-34}} \approx 25.61 \times 10^{73} \approx 2.6 \times 10^{74} \]

Hence, the quantum number that characterizes Earth's revolution around the Sun is:

\( \boxed{2.6 \times 10^{74}} \)

Answer 2

According to Bohr's model quantization condition, the angular momentum (L) of an object revolving in a stable orbit is an integer multiple of \( \frac{h}{2\pi} \), where \( h \) is Planck's constant and \( n \) is the principal quantum number. \[ L = n \frac{h}{2\pi} \] The angular momentum \( L \) for an object of mass \( m \) moving with speed \( v \) in an orbit of radius \( r \) is given by: \[ L = mvr \] Combining these two equations, we get: \[ mvr = n \frac{h}{2\pi} \]

We are asked to find the quantum number \( n \) that characterizes the Earth's revolution around the Sun. We are given the following values:

• Mass of the Earth, \( m = 6 \times 10^{24} \) kg
• Radius of Earth's orbit, \( r = 1.5 \times 10^{11} \) m
• Orbital speed of Earth, \( v = 3 \times 10^{4} \) m/s
• Planck's constant, \( h = 6.626 \times 10^{-34} \) J·s

 

Rearranging the quantization condition to solve for \( n \): \[ n = \frac{mvr}{\left(\frac{h}{2\pi}\right)} = \frac{2\pi mvr}{h} \]

First, let's calculate the angular momentum \( L = mvr \): \[ L = (6 \times 10^{24} \text{ kg}) \times (3 \times 10^{4} \text{ m/s}) \times (1.5 \times 10^{11} \text{ m}) \] \[ L = (6 \times 3 \times 1.5) \times 10^{(24 + 4 + 11)} \text{ kg m}^2/\text{s} \] \[ L = 27 \times 10^{39} \text{ Js} \]

Now, substitute the value of \( L \) and \( h \) into the equation for \( n \): \[ n = \frac{2\pi \times (27 \times 10^{39} \text{ Js})}{6.626 \times 10^{-34} \text{ Js}} \] Using \( \pi \approx 3.14159 \): \[ n = \frac{2 \times 3.14159 \times 27}{6.626} \times 10^{39 - (-34)} \] \[ n = \frac{169.64586}{6.626} \times 10^{73} \] \[ n \approx 25.603 \times 10^{73} \] Writing this in standard scientific notation: \[ \mathbf{n \approx 2.5603 \times 10^{74}} \]

Comparing this result with the given options:

• 8.57 x 10⁶⁴
• 5.98 x 10⁶⁴
• 2.57 x 10⁷⁴
• 2.57 x 10³⁸

The calculated value \( 2.5603 \times 10^{74} \) is closest to \( 2.57 \times 10^{74} \).

 

The quantum number characterizing the Earth's revolution is approximately 2.57 x 10⁷⁴.