The time of revolution of an electron around a nucleus of charge $Ze$ in the $n^{th}$ Bohr orbit is directly proportional to

To calculate the size of a hydrogen ion $(H^-)$ using the Bohr's model,we assume that its two electrons move in an orbit such that they are always on diametrically opposite sides of the nucleus. With each electron having the angular momentum $\hbar = h / 2\pi$ and taking electron interaction into account,the radius of the orbit in terms of the Bohr's radius of hydrogen atom $a_B = \frac{4\pi\varepsilon_0\hbar^2}{me^2}$ is

If an electron is revolving in its Bohr orbit having Bohr radius of $0.529 Å$,then the radius of the third orbit is

$A$ $\mu$-meson of charge '$e$',mass $208 m_e$ moves in a circular orbit around a heavy nucleus having charge $+3e$. The quantum state '$n$' for which the radius of the orbit is the same as that of the first Bohr orbit for a hydrogen atom is [approximately]:

An excited $He^+$ ion emits two photons in succession, with wavelengths $108.5 \, nm$ and $30.4 \, nm$, in making a transition to the ground state. The quantum number $n$, corresponding to its initial excited state is $n = ........$ (for a photon of wavelength $\lambda$, energy $E = \frac{1240 \, eV}{\lambda \, (in \, nm)}$).

If in a hydrogen atom,the radius of the ${n^{th}}$ Bohr orbit is ${r_n}$ and the frequency of revolution of the electron in the ${n^{th}}$ orbit is ${f_n}$,choose the correct option.