In a hydrogen atom,the electron makes a transition from the $(n+1)^{\text{th}}$ level to the $n^{\text{th}}$ level. If $n >> 1$,the frequency of the radiation emitted is proportional to:

$A$ sample of hydrogen atoms is in an excited state (all the atoms). The photons emitted from this sample are made to pass through a filter through which light having a wavelength greater than $800 \ nm$ can only pass. Only one type of photon is found to pass through the filter. The sample's initial excited state is: [Take $hc = 1240 \ eV \cdot nm$,ground state energy of hydrogen atom = $-13.6 \ eV$.]

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.....$eV$.

An electron jumps from the $5^{th}$ orbit to the $4^{th}$ orbit of a hydrogen atom. Taking the Rydberg constant as $10^7 \ m^{-1}$,what will be the frequency of the radiation emitted?

The energy of an electron in the first excited state of an $H$-atom is $-3.4 \ eV$. Its kinetic energy is ........ $eV$.

If the binding energy of a ground state electron in a hydrogen atom is $13.6\,eV$,then the energy required to remove the electron from the second excited state of $Li^{2+}$ will be $x \times 10^{-1}\,eV$. The value of $x$ is $...........$