The angular momentum of an electron in a hydrogen atom is proportional to: (Where $r$ is the radius of the orbit of the electron)

The de-Broglie wavelength of the electron in the ground state of a hydrogen atom is

$A$ particular hydrogen-like ion emits radiation of frequency $2.92 \times 10^{15} \text{ Hz}$ when it makes a transition from $n=3$ to $n=1$. The frequency in $\text{Hz}$ of radiation emitted in the transition from $n=2$ to $n=1$ will be: (in $\times 10^{15}$)

The ratio of the total energy of the $2^{\text{nd}}$ orbit electron for the hydrogen atom $(_1H^1)$ to that of the helium ion $(He^+)$ $(_2^4He)$ is:

$A$ hydrogen atom changes its state from $n=3$ to $n=2$. Due to recoil,the percentage change in the wavelength of emitted light is approximately $1 \times 10^{-n}$. The value of $n$ is . . . . . . .[Given $Rhc=13.6 \text{ eV}, hc=1242 \text{ eV nm}, h=6.6 \times 10^{-34} \text{ J s}$,mass of the hydrogen atom $=1.67 \times 10^{-27} \text{ kg}$]

The momentum of an electron revolving in the $n^{\text{th}}$ orbit is given by: (Symbols have their usual meanings)