Monochromatic radiation is incident on a hydrogen $(H)$ sample which is in the ground state. If the hydrogen atoms emit radiation of $10$ different wavelengths after absorbing the incident radiation, then the wavelength of the incident radiation is (Let $hc = 1242 \text{ eV-nm}$) (in $\text{ nm}$)

Consider the $3^{rd}$ orbit of $He^{+}$ (Helium) using a non-relativistic approach. The speed of the electron in this orbit will be (given $K = 9 \times 10^9 \; N \cdot m^2/C^2$,$Z = 2$,and $h = 6.6 \times 10^{-34} \; J \cdot s$).

The electron in a hydrogen atom makes a transition $n_1 \rightarrow n_2$,where $n_1$ and $n_2$ are the principal quantum numbers of the two states. Assume the Bohr model to be valid. The frequency of orbital motion of the electron in the initial state is $1/27$ of that in the final state. The possible values of $n_1$ and $n_2$ are

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:

According to the de Broglie hypothesis,if the de Broglie wavelength of an electron in a hydrogen atom orbit of radius $5.3 \times 10^{-11} \ m$ is $10^{-10} \ m$,then the principal quantum number of the electron is ..........

Consider an electron in a hydrogen atom,revolving in its second excited state (having radius $4.65 \, \mathring{A}$). The de-Broglie wavelength of this electron is .... $\mathring{A}$.