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:

The radius of the first orbit of hydrogen is $r_{H}$,and the energy in the ground state is $-13.6 \text{ eV}$. Considering a $\mu^{-}$-particle with a mass $207 m_e$ revolving around a proton as in a hydrogen atom,the energy and radius of the proton and $\mu^{-}$-combination respectively in the first orbit are (assume the nucleus to be stationary):

The magnitude of acceleration of the electron in the $n$th orbit of a hydrogen atom is $a_{H}$ and that of a singly ionised helium atom is $a_{He}$. The ratio of $a_{H} : a_{He}$ 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 $=1.67 \times 10^{-27} \text{ kg}$]

Assuming the atom in the ground state,the expression for the magnetic field at a point (nucleus) in a hydrogen atom due to the circular motion of the electron is [$\mu_{0} =$ permeability of free space,$\epsilon_{0} =$ permittivity of free space,$m =$ mass of electron,$e =$ electronic charge,$h =$ Planck's constant].

When an electron jumps from the orbit $n=2$ to $n=4$,the wavelength of the radiation absorbed will be ($R$ is Rydberg's constant).