Q.In accordance with the Bohr's model, find the quantum number that characterises the earth's revolution around the sun in an orbit of radius $1.5 \times 10^{11}\ \text{m}$ with orbital speed $3 \times 10^{4}\ \text{m/s}$. (Mass of earth $= 6.0 \times 10^{24}\ \text{kg}$.)

Concept: Bohr Model Quantization — Bohr proposed that angular momentum in allowed orbits is an integer multiple of $\frac{h}{2\pi}$.

Reasoning:

1. The angular momentum of Earth in its orbit is $L = m v r$.
2. Bohr’s quantization condition: $L = n \frac{h}{2\pi}$, where $n$ is the quantum number.
3. Equate and solve for $n$:

Calculation:

$n = \frac{2\pi \times (6.0 \times 10^{24}) \times (3 \times 10^{4}) \times (1.5 \times 10^{11})}{6.63 \times 10^{-34}}$ …

The Bohr model was originally proposed for the hydrogen atom, where the electron’s angular momentum around the nucleus is quantized in integer multiples of $\hbar = h/2\pi$. The key insight is that this quantization condition — $mvr = n\hbar$ — is not limited to atoms. It can be applied to any orbiting system, including Earth around the Sun. The result tells us how “quantum” the orbit is: a small $n$ means the system is truly quantum, while a huge $n$ (like here) means the orbit behaves classically, because the spacing between adjacent quantum levels becomes vanishingly small.

Let’s work through the numbers step by step.

1. Write down the quantization condition. Bohr’s postulate for angular momentum is:

where $m$ is the mass of the orbiting body (Earth), $v$ is its orbital speed, $r$ is the orbital radius, $n$ is the quantum number (an integer), and $\hbar = \frac{h}{2\pi}$ with $h = 6.626 \times 10^{-34}\ \text{J·s}$.

2. Identify the given values.

   • $r = 1.5 \times 10^{11}\ \text{m}$
   • $v = 3 \times 10^{4}\ \text{m/s}$
   • $m = 6.0 \times 10^{24}\ \text{kg}$
   • $h = 6.626 \times 10^{-34}\ \text{J·s}$, so $\hbar = \frac{h}{2\pi} \approx 1.0546 \times 10^{-34}\ \text{J·s}$

3. Calculate the left-hand side: Earth’s angular momentum.

Multiply stepwise:

$6.0 \times 3 = 18$, and $10^{24} \times 10^{4} = 10^{28}$, so $m v = 18 \times 10^{28} = 1.8 \times 10^{29}\ \text{kg·m/s}$.

Then $L = (1.8 \times 10^{29}) \times (1.5 \times 10^{11}) = 2.7 \times 10^{40}\ \text{kg·m}^2/\text{s}$.

4. Solve for $n$. From $L = n \hbar$, we have: …

Method: Angular Momentum Quantization (Bohr's Postulate)

Bohr's model says that stable orbits are those where the angular momentum of the revolving body is an integer multiple of $\frac{h}{2\pi}$.

Step 1: Write the quantization condition

For any orbiting body in Bohr's model:

where $m$ is mass, $v$ is orbital speed, $r$ is orbit radius, $h$ is Planck's constant, and $n$ is the quantum number (a positive integer).

Step 2: Identify the given values

• $m = 6.0 \times 10^{24}\ \text{kg}$
• $v = 3 \times 10^{4}\ \text{m/s}$
• $r = 1.5 \times 10^{11}\ \text{m}$
• $h = 6.63 \times 10^{-34}\ \text{J·s}$ (standard value)

Step 3: Solve for $n$

Rearrange the quantization condition:

Step 4: Substitute and compute

First compute the numerator:

Multiply stepwise:

• $6.0 \times 10^{24} \times 3 \times 10^{4} = 18 \times 10^{28}$
• $18 \times 10^{28} \times 1.5 \times 10^{11} = 27 \times 10^{39}$ …

Common Mistakes in the Bohr Model Quantization Problem

This question asks you to treat the Earth-Sun system as if it obeyed Bohr's angular momentum quantization condition — a purely hypothetical exercise, since Bohr's model applies to atoms, not planetary orbits. The calculation itself is straightforward, but students make several predictable errors.

Mistake 1: Forgetting the $n$ must be an integer

The most fundamental error is to compute a value for $n$ and not check whether it makes physical sense. The Bohr condition is:

Plugging in the numbers:

The numerator is roughly $1.7 \times 10^{41}$, and dividing by $6.63 \times 10^{-34}$ gives $n \approx 2.6 \times 10^{74}$.

How to avoid: Always note that $n$ must be a positive integer. Here it is an astronomically large integer — that's fine conceptually, but if your calculation gave a non-integer like $2.6 \times 10^{74}$, you've made an arithmetic error. The actual value is indeed an integer (approximately $2.55 \times 10^{74}$), but the point is that Bohr quantization is meaningless for macroscopic orbits.

Mistake 2: Using the wrong value of Planck's constant

Students sometimes use $h = 6.63 \times 10^{-34}$ correctly but then accidentally use $h = 6.63 \times 10^{-34}$ in the denominator when the formula has $h/2\pi$. Or they use $\hbar = h/2\pi$ but forget the factor of $2\pi$ entirely.

How to avoid: Write the formula explicitly before substituting numbers. Either use $mvr = n \frac{h}{2\pi}$ or $mvr = n \hbar$ — but be consistent.

Mistake 3: Unit mismatch or missing conversions

All quantities must be in SI units. The radius is given in metres, speed in m/s, mass in kg — so no conversion is needed here. But students sometimes treat the radius as $1.5 \times 10^{11}$ cm or the speed as $3 \times 10^{4}$ km/h without converting.

How to avoid: Before substituting, quickly verify each quantity is in SI base units. If any is not, convert first.

Mistake 4: Arithmetic errors with large exponents …