According to de Broglie,if the wavelength of an electron in the second orbit is $10^{-9} \ m$,then the circumference of the orbit is .......

When a hydrogen atom emits a photon of energy $12.09 \,eV$,its orbital angular momentum changes by (where $h$ is Planck's constant)

The magnetic moment $(\mu)$ of an electron revolving around the nucleus varies with the principal quantum number $n$ as

In $Li^{++}$ ion,an electron in the first Bohr orbit is excited to a higher energy level by a radiation of wavelength $\lambda$. When the ion de-excites to the ground state in all possible ways (including intermediate emissions),a total of six spectral lines are observed. What is the value of $\lambda$ in $nm$? (Given: $h = 6.63 \times 10^{-34} \, J \cdot s, c = 3 \times 10^8 \, m/s, 1 \, eV = 1.6 \times 10^{-19} \, J$)

$A$ free hydrogen atom after absorbing a photon of wavelength $\lambda_{a}$ gets excited from the state $n=1$ to the state $n=4$. Immediately after that,the electron jumps to $n=m$ state by emitting a photon of wavelength $\lambda_{e}$. Let the change in momentum of the atom due to the absorption and the emission be $\Delta p_{a}$ and $\Delta p_{e}$,respectively. If $\lambda_{a} / \lambda_{e} = 1/5$,which of the following options is/are correct?
[Use $hc = 1242 \text{ eV nm}$; $1 \text{ nm} = 10^{-9} \text{ m}$,$h$ and $c$ are Planck's constant and speed of light,respectively]
$(1)$ $\lambda_{e} = 418 \text{ nm}$
$(2)$ The ratio of kinetic energy of the electron in the state $n=m$ to the state $n=1$ is $1/4$
$(3)$ $m=2$
$(4)$ $\Delta p_{a} / \Delta p_{e} = 1/2$

When an electron in a hydrogen atom jumps from the third excited state to the ground state,the de-Broglie wavelength associated with the electron becomes