An electron in the hydrogen atom jumps from the first excited state to the ground state. What will be the percentage change in the speed of the electron (in $\%$)?

Angular momentum of an electron in a hydrogen atom is $\frac{3h}{2\pi}$ ($h$ is the Planck's constant). The kinetic energy $(KE)$ of the electron is (in $\text{ eV}$)

What will be the angular momentum of an electron,if the energy of this electron in an $H$-atom is $-1.5 \, eV$ (in $J \cdot s$)?

Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr model to this atom. The longest wavelength photon that will be emitted has wavelength $\lambda$ (given in terms of the Rydberg constant $R$ for the hydrogen atom) equal to:

In a hydrogen atom,the radius of the $n^{th}$ Bohr orbit is $r_n$. The graph between $\log \left( \frac{r_n}{r_1} \right)$ and $\log n$ will be:

In the Bohr's hydrogen atom model,the radius of the stationary orbit is directly proportional to ($n =$ principle quantum number)