The de-Broglie wavelength of the electron in the ground state of the hydrogen atom is ..... (radius of the first orbit of hydrogen atom $= 0.53 \ \text{Å}$). (in $\text{Å}$)

$A$ 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 energy of a hydrogen atom in the ground state is $-13.6 \, eV$. The energy of the $He^+$ ion in the first excited state will be .... $eV$.

The magnetic moment of an orbital electron in the first orbit of an $\text{H}$-atom is $\mu_{B}$. What is the magnetic moment of that electron in the second orbit?

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

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: