$A$ $12.5 \ eV$ electron beam is used to bombard gaseous hydrogen at room temperature. It will emit:

The ground state energy of a hydrogen atom is $-13.6 \text{ eV}$. When its electron is in the first excited state,its excitation energy is:

In the $n^{th}$ orbit,the energy of an electron is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$ for a hydrogen atom. The energy required to excite the electron from the first orbit $(n=1)$ to the second orbit $(n=2)$ will be..... eV.

If $\lambda_1$ and $\lambda_2$ are the wavelengths of the photons emitted when electrons in the $n^{\text{th}}$ orbit of a hydrogen atom fall to the first excited state and ground state respectively,then the value of $n$ is:

The binding energy of the electron in a hydrogen atom is $13.6\, eV$. The energy required to remove the electron from the first excited state of $Li^{++}$ is ......... $eV$.

$A$ hydrogen atom is bombarded with electrons accelerated through a potential difference of $V$, which causes excitation of hydrogen atoms. If the experiment is performed at $T = 0 \,K$, the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be $\frac{\alpha}{10} \,V$, where $\alpha = $ . . . . . . .