The de-Broglie wavelength $(\lambda_B)$ associated with the electron orbiting in the second excited state of a hydrogen atom is related to that in the ground state $(\lambda_G)$ by:

If $E$ and $L$ denote the magnitude of total energy and angular momentum of a revolving electron in the $n^{th}$ Bohr orbit, then:

The figure shows standing de-Broglie waves due to the revolution of an electron in a certain orbit of a hydrogen atom. Then,the expression for the orbit radius is (All notations have their usual meanings).

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 $E = \frac{1240 \, eV}{\lambda \, (in \, nm)}$).

An electron has a mass of $9.1 \times 10^{-31} \ kg$. It revolves around the nucleus in a circular orbit of radius $0.529 \times 10^{-10} \ m$ at a speed of $2.2 \times 10^6 \ m/s$. The magnitude of its linear momentum in this motion is:

What is the ratio of the time periods of an electron in the $n = 2$ and $n = 1$ orbits of a hydrogen atom (in $:1$)?