When an electron orbiting in a hydrogen atom in its ground state jumps to a higher excited state, the de-Broglie wavelength associated with it

A hydrogen atom, initially in the ground state, is excited by absorbing a photon of wavelength $980 \ \text{\AA}$. The radius of the atom in the excited state, in terms of Bohr radius $a_0$, will be (given $hc = 12500 \ \text{eV-\AA}$). (in $a_0$)

$A$ hydrogen atom,initially at rest in its ground state,absorbs a photon of frequency $v_1$ and ejects the electron with a kinetic energy of $10 eV$. The electron then combines with a positron at rest to form a positronium atom in its ground state and simultaneously emits a photon of frequency $v_2$. The center of mass of the resulting positronium atom moves with a kinetic energy of $5 eV$. It is given that the positron has the same mass as that of the electron and the positronium atom can be considered as a Bohr atom,in which the electron and the positron orbit around their center of mass. Considering no other energy loss during the whole process,the difference between the two photon energies (in $eV$) is $....$ (in $.80$)

Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model?

The angular momentum of an electron in the hydrogen atom is $\frac{3h}{2\pi}$. Here $h$ is Planck's constant. The kinetic energy of this electron is.....$eV$

An electron in a hydrogen atom makes a transition from $n_1$ to $n_2$. If the time period of the electron in the initial state is eight times that in the final state,then: