The ionisation potential of a hydrogen atom is $13.6 \, V$. The energy required to remove an electron in the $n = 2$ state of the hydrogen atom is.....$eV$.

What is the total energy of an electron at an infinite distance from the nucleus in an atom?

$A$ 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 $n^{th}$ 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.

$A$ photon of wavelength $\lambda$ is absorbed by an electron confined to a box of length $L = \sqrt{(35 h \lambda / 8 m c)}$. As a result,the electron makes a transition from state $k=1$ to the state $n$. Subsequently,the electron transitions from the state $n$ to the state $m$ by emitting a photon of wavelength $\lambda^{\prime} = 1.75 \lambda$. Then,

Due to transitions among its first three energy levels,a hydrogenic atom emits radiation at three discrete wavelengths $\lambda_1, \lambda_2$ and $\lambda_3$ (where $\lambda_1 < \lambda_2 < \lambda_3$). Then,