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To start with its a homework problem, quite lengthy. 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 BOHRs model of the atom is applicable to this system

  1. Derive an expression for radius of nth 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 mass of the particle revolving = M

$Speed = v$

Also $M=208m_e$

Electrostatic force is the centripetal force. Therefore

$$\frac{Mv^2}{r}=\frac{(ke)(3e)}{r^2}$$ $$v^2=\frac{3ke^2}{208m_er}$$ From Bohrs model, $$m_evr=\frac{nh}{2\pi}$$ where h=planck's constant

Therefore, $$v=\frac{nh}{2\pi m_er}$$ 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, $$\frac{n^2h^2}{4(\pi)^2(m_e)^2r^2}=\frac{3ke^2}{208m_er}$$ After solving for r, we get something like this, $$r=\frac{n^2h^2}{4(\pi)^23ke^2208m_e}$$ Now this is correct. The problem is in the second and third parts. Now when I put $r=.53*10^{-10}$m I do NOT get the required answer.

And to approach the third part, I started with the standard rydberg equation, $$\frac{1}{\lambda}=RZ^2[\frac{1}{(n_f)^2}-\frac{1}{(n_i)^2}]$$ I plugged each value, $n_i=3$,$n_f=1$$,Z=3$ ; but again didn't get the answer correct. Please DON'T make edits on this question.

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

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2 Answers 2

up vote 3 down vote accepted

To answer the second part:

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.

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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 $$\frac{208m_ev^2}{r}=K\frac{3e^2}{r^2}\wedge \color{red}{208}m_evr=\frac{nh}{2\pi} $$

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2  
but angular momentum should be $Mvr$ AFAIK –  eaxdpiotnyeantial Aug 7 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 Aug 7 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 at 14:22

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