Kinetic energy of an electron in one of the orbits of a hydrogen atom is $x$. Then, its total energy is . . . . . . .

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 $a_0$)

The triply ionized beryllium $(Be^{3+})$ has the same electron orbital radius as that of the ground state of hydrogen. Hence, the energy state of triply ionized beryllium is (Given $Z = 4$ for beryllium)

In a hydrogen atom,an electron makes a transition from the $5^{\text{th}}$ orbit to the $3^{\text{rd}}$ orbit. The change in the angular momentum for this electron is . . . . . . .

Using the formula for the radius of the $n^{th}$ orbit $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$,derive an expression for the total energy of an electron in the $n^{th}$ Bohr orbit.

For principal quantum number $n = 3$,the possible values of orbital quantum number $l$ are