The first excited state of a hydrogen atom is $10.2 \, eV$ above its ground state. The temperature needed to excite hydrogen atoms to the first excited level is

The kinetic energy of an electron in the $n^{th}$ orbit of a hydrogen-like species with atomic number $Z$ is $13.6 \frac{Z^2}{n^2} \ eV$. The potential energy of this electron in the same orbit will be:

An electron collides with a fixed hydrogen atom in its ground state. The hydrogen atom gets excited and the colliding electron loses all its kinetic energy. Consequently,the hydrogen atom may emit a photon corresponding to the largest wavelength of the Balmer series. The minimum kinetic energy $(K.E.)$ of the colliding electron will be.....$eV$.

The energy (in $eV$) required to excite an electron from $n=2$ to $n=4$ state in a hydrogen atom is:

$A$ singly ionised helium atom in an excited state $(n=4)$ emits a photon of energy $2.6 \,eV$. Given that the ground state energy of hydrogen atom is $-13.6 \,eV$,the energy $E_f$ and quantum number $n$ of the resulting state are respectively,

In the hydrogen atom,if the reference level of potential energy is assumed to be zero at the ground state level,choose the incorrect statement.