The de-Broglie wavelength of the electron in the ground state of a hydrogen atom is ........ $\mathring{A}$.

$A$ hydrogen atom in the ground state absorbs $\Delta E$ amount of energy. If the orbital angular momentum of the electron is increased by $\frac{h}{2 \pi}$ ($h=$ Planck constant), then the magnitude of $\Delta E$ is (in $eV$)

As the electron in a Bohr orbit of a Hydrogen atom transitions from state $n = 2$ to $n = 1$,how do the kinetic energy $K$ and potential energy $U$ change?

$A$ light of energy $12.75 \; eV$ is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is $\frac{x}{\pi} \times 10^{-17} \; eVs$. The value of $x$ is $........$ (use $h=4.14 \times 10^{-15} \; eVs$)

When an electron in a hydrogen atom is excited from its $4^{th}$ to $5^{th}$ stationary orbit,the change in angular momentum of the electron is (Planck's constant: $h = 6.6 \times 10^{-34} \ J \cdot s$)

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