Q.(a) Using the Bohr's model calculate the speed of the electron in a hydrogen atom in the $n = 1, 2,$ and $3$ levels.

Concept: Bohr Model Quantization — the electron's angular momentum is quantised, $m_e v_n r_n = n\dfrac{h}{2\pi}$, and the Coulomb force supplies the centripetal force.

1. Speed of the electron Solving the force-balance and quantisation equations together gives $v_n = \dfrac{e^2}{2\varepsilon_0 n h}$, which falls as $1/n$. Substituting the constants gives $v_1 \approx 2.19 \times 10^6\ \text{m/s}$, so:

   $v_1 \approx 2.19 \times 10^6\ \text{m/s}, \quad v_2 = \frac{v_1}{2} \approx 1.09 \times 10^6\ \text{m/s}, \quad v_3 = \frac{v_1}{3} \approx 7.29 \times 10^5\ \text{m/s}$

2. Orbital period Using $r_n = n^2 a_0$ (with $a_0 \approx 5.29\times10^{-11}\ \text{m}$) and $T_n = \dfrac{2\pi r_n}{v_n}$: …

Why Bohr's model works for this

Bohr's model combines classical circular motion with one quantum condition: the electron's angular momentum is an integer multiple of $\frac{h}{2\pi}$. The Coulomb force provides the centripetal force, and quantising the angular momentum ties the speed $v$ to the orbit radius $r$ — solving the two together gives both in terms of $n$ alone.

Step-by-step calculation

1. The two governing equations

For an electron of mass $m_e$ and charge $-e$ orbiting a proton (charge $+e$) in a circular orbit of radius $r$ with speed $v$:

• Coulomb force = centripetal force:

• Bohr's quantisation of angular momentum:

2. Solve for the speed $v_n$

Eliminating $r$ between these two equations gives:

The speed falls as $1/n$ — higher orbits mean slower electrons.

3. Substitute the constants

Using $e = 1.602 \times 10^{-19}\ \text{C}$, $\varepsilon_0 = 8.854 \times 10^{-12}\ \text{F/m}$, $h = 6.626 \times 10^{-34}\ \text{J·s}$:

Since $v_n = v_1 / n$:

• $v_1 \approx 2.19 \times 10^6\ \text{m/s}$
• $v_2 = v_1/2 \approx 1.09 \times 10^6\ \text{m/s}$
• $v_3 = v_1/3 \approx 7.29 \times 10^5\ \text{m/s}$

4. Find the orbital radius $r_n$

From the two governing equations, $r_n = n^2 a_0$, where $a_0 = \frac{\varepsilon_0 h^2}{\pi m_e e^2} \approx 5.29 \times 10^{-11}\ \text{m}$ is the Bohr radius:

• $r_1 = 5.29 \times 10^{-11}\ \text{m}$ …

Method: Bohr's Quantization of Angular Momentum

This problem uses the Bohr quantization condition — the idea that angular momentum comes in discrete packets — combined with the Coulomb force providing the centripetal acceleration.

Step 1: Write the two governing equations

Quantization of angular momentum (Bohr's postulate):

Coulomb force = centripetal force (for a hydrogen nucleus with charge $+e$):

Where:

• $m_e = 9.11 \times 10^{-31}\ \text{kg}$
• $e = 1.60 \times 10^{-19}\ \text{C}$
• $h = 6.63 \times 10^{-34}\ \text{J·s}$
• $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/\text{N·m}^2$

Step 2: Solve for speed $v$ in terms of $n$

From the quantization condition: $r = \frac{n h}{2\pi m_e v}$

Substitute into the force equation:

This simplifies to:

Cancel $v^2$ (non-zero) and rearrange:

Cancel $\pi$ and $m_e$:

Multiply both sides by $n h$:

This is the speed of the electron in the $n$th Bohr orbit.

Step 3: Calculate $v_1$, $v_2$, $v_3$

First compute the constant factor:

Numerator: $2.56 \times 10^{-38}$

Denominator: $2 \times 8.85\times10^{-12} \times 6.63\times10^{-34} = 1.173 \times 10^{-44}$

So the constant $= \frac{2.56\times10^{-38}}{1.173\times10^{-44}} = 2.18 \times 10^6\ \text{m/s}$

Therefore:

Part (b): Orbital period

Method: Period $T = \frac{\text{circumference}}{\text{speed}} = \frac{2\pi r}{v}$

We need $r$ for each $n$. From the quantization condition:

This is the Bohr radius $a_0 = 5.29 \times 10^{-11}\ \text{m}$ when $n=1$. …

Common Mistakes in Bohr Model Calculations

Students often lose marks on this exact problem because they rush through the algebra or misapply the quantization condition. Let me walk through the most frequent errors and how to fix each.

Mistake 1: Using the wrong formula for velocity

Many students try to derive velocity from $mvr = \frac{nh}{2\pi}$ alone, forgetting that the Coulomb force provides the centripetal force. They end up with an expression that still contains $r$, which they don't know yet.

How to avoid: Always start from the force balance equation:

This gives $v^2 = \frac{ke^2}{mr}$. Then combine with the quantization condition $mvr = n\hbar$ (where $\hbar = h/2\pi$) to eliminate $r$. You get:

This is the clean, direct formula. Memorise it — it saves time and prevents algebra errors.

Mistake 2: Plugging in numbers with inconsistent units

Students use $k = 9 \times 10^9$ (SI), $e = 1.6 \times 10^{-19}$ C, but then use $h = 6.63 \times 10^{-34}$ J·s — all correct — but forget that $\hbar = h/2\pi$, not $h$ itself. This off-by-a-factor-of-$2\pi$ error is extremely common.

How to avoid: Write $\hbar$ explicitly as $h/2\pi$ in your formula before substituting numbers. For $n=1$:

Now substitute: $k = 9 \times 10^9$, $e = 1.6 \times 10^{-19}$, $h = 6.63 \times 10^{-34}$.

Calculate stepwise: $e^2 = 2.56 \times 10^{-38}$, so numerator = $2\pi \times 9 \times 10^9 \times 2.56 \times 10^{-38} = 2\pi \times 2.304 \times 10^{-28} \approx 1.447 \times 10^{-27}$. Divide by $6.63 \times 10^{-34}$ to get $v_1 \approx 2.18 \times 10^6$ m/s.

Mistake 3: Forgetting that $v \propto 1/n$

Once you have $v_1$, students sometimes recalculate everything from scratch for $n=2$ and $n=3$, wasting time and inviting arithmetic errors.

How to avoid: From the formula $v_n = \frac{ke^2}{n\hbar}$, it's clear that $v_n = v_1 / n$. So:

• $v_2 = v_1 / 2 \approx 1.09 \times 10^6$ m/s
• $v_3 = v_1 / 3 \approx 7.27 \times 10^5$ m/s

No need to redo the full substitution.

Mistake 4: Confusing orbital period with frequency

For part (b), students often write $T = \frac{2\pi r}{v}$ but then use the wrong $r$ or forget that $r$ also depends on $n$.

How to avoid: First, recall that $r_n = n^2 a_0$, where $a_0 = \frac{\hbar^2}{mke^2} \approx 5.29 \times 10^{-11}$ m is the Bohr radius. Then:

So $T_n \propto n^3$. Calculate $T_1$ once, then multiply by $n^3$ for higher levels.

For $n=1$:

$T_1 = \frac{2\pi (5.29 \times 10^{-11})}{2.18 \times 10^6} \approx 1.52 \times 10^{-16} \text{ s}$ …