# Finding orbit radius using the Bohr model and Rydberg equation

A particle of mass equal to 208 times the mass of an electron moves in a circular orbit around a nucleus of charge $+3e$. Assuming the Bohr model of the atom is applicable to this system,

1. Derive an expression for radius of $n$th Bohr orbit.
2. Find the value of $n$ for which the radius is equal to the radii of first orbit of hydrogen.
3. Find the wavelength of radiation emitted when revolving particle jumps from third orbit to the first.

Now, I did the first part and got the answer correct. Here's what I did.

Suppose the mass of the particle revolving is $M$, its speed is $v$, and $M = 208 m_{e}$. Electrostatic force is the centripetal force. Therefore

\begin{align} \frac{Mv^2}{r} &= \frac{(ke)(3e)}{r^2} \\ v^2 &= \frac{3ke^2}{208m_{e}r} \end{align}

From the Bohr model,

$$m_{e}vr=\frac{nh}{2\pi}$$

where $h$ is Planck's constant. Therefore,

$$v = \frac{nh}{2\pi m_{e}r}$$

Squaring it,

$$v^2 = \frac{n^2h^2}{4(\pi)^2(m_e)^2r^2}$$

Equating the two equations which have $v^2$ in them,

$$\frac{n^2h^2}{4(\pi)^2(m_e)^2r^2} = \frac{3ke^2}{208m_{e}r}$$

After solving for $r$, we get something like this,

$$r = \frac{n^2h^2}{4(\pi)^{2}3ke^{2}208m_e}$$

All of the above is correct. The problem is in the second and third parts; when I put $r =\pu{0.53*10^{-10} m}$ I do NOT get the required answer. To approach the third part, I started with the standard Rydberg equation,

$$\frac{1}{\lambda} = \mathcal{R}Z^2 \left( \frac{1}{n_f^2}-\frac{1}{n_i^2} \right)$$

I plugged in each value, $n_i = 3, n_f = 1, Z = 3$; but again didn't get the answer correct.

The answer to the second part is 25 $(n=25)$; and to the third is 55.2 picometers.

We know $M=208m_e$, $Z=3$, $\hbar=\frac{h}{2\pi}$.

Part one has a mistake, as it is

\begin{align} &&\frac{Mv^2}{r}&=\mathcal{k_e}\cdot\frac{(\mathcal{e})(Z\mathcal{e})}{r^2}\\ &&Mvr&=n\hbar\\ \implies&& r &=\frac{n^2\hbar^2}{M\cdot\mathcal{k_e}\cdot Z \mathcal{e}^2} \end{align}

We also know the Bohr radius:

$$a_0= \mathcal{k_e}\cdot\frac{\hbar^2}{m_e\cdot\mathcal{e}^2} \approx 5{,}29 \cdot 10^{-11}\mathrm{m}$$

Therefore we can write and cancel:

\begin{align} && r&=a_0\\ && \frac{\color{\green}{\hbar^2}}{\color{\red}{\mathcal{k_e}}\cdot m_e\cdot\color{\navy}{\mathcal{e}^2}} &=\frac{n^2\color{\green}{\hbar^2}}{M\cdot\color{\red}{\mathcal{k_e}}\cdot Z \color{\navy}{\mathcal{e}^2}}\\ \therefore&& Z\frac{M}{m_e}&=n^2\\ \therefore&& n&=\sqrt{Z\cdot208}\approx25 \end{align}

The third part:

The Rydberg Formula is given as

$$\frac{1}{\lambda_{\mathrm{vac}}} = \mathcal{R}Z^2 \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$

with the Rydberg $\mathcal{R}$ constant defined for a photon emitted by an electron. We can safely assume, that the mass of the nucleus equals to three protons since it has a charge of 3: $T=Z\cdot m_p$. Taking into account that $m_p\approx 1836m_e$, we arrive at

$$\mathcal{R} =\frac{\mathcal{R}_\infty}{1+\frac{M}{T}} = \frac{\mathcal{R}_\infty}{1+\frac{208m_e}{3\cdot1836m_e}}$$

Now the Rydberg constant has to be modified to include the mass of the particle:

$$\mathcal{R}_\infty = \frac{M e^4}{8 c \varepsilon_0^2 h^3}=208\mathcal{R}_{m_e,\infty}$$

With $\mathcal{R}_{m_e,\infty} = 1.097\cdot 10^7~\mathrm{m^{-1}}$ (wikipedia), I got to $\lambda_\mathrm{vac}=56.8~\mathrm{pm}$.

Without taking the reduced mass into account, i.e. $\mathcal{R}\approx\mathcal{R}_\infty$ I arrived at $\lambda_\mathrm{vac}=54.8~\mathrm{pm}$.

Both values I would consider reasonably close to the solution given.

A particle of mass equal to $208$ times the mass of electron moves in a circular orbit around a nucleus of charge $+3e$. Assuming the Bohr's model of the atom is applicable to this system:

• Derive an expression for radius of nth bohr orbit

• Find the value of n for which the radius is equal to the radii of first orbit of hydrogen.

• Find the wavelength of radiation emitted when revolving particle jumps from third orbit to the first.

You need to correct this For first part $$\frac{208m_ev^2}{r}=K\frac{3e^2}{r^2}\wedge \color{red}{208}m_evr=\frac{nh}{2\pi}$$

• but angular momentum should be $Mvr$ AFAIK – RE60K Aug 7 '14 at 4:12
• You are right, and if you look at the solution where @Hardik arrived, this was actually taken into account. Though the mathematical part of the deviation is incorrect. – Martin - マーチン Aug 7 '14 at 6:05
• I think this is helpful enough as a "nudge" to get the OP moving along, but something like this should probably be a comment in the future (or could certainly be fleshed out a bit into an answer with more information included). – jonsca Aug 7 '14 at 14:22