To start with its a homework problem, quite lengthy.
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,
- Derive an expression for radius of $n$th 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.
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.