Let $T_{1}$ and $T_{2}$ be the energy of an electron in the first and second excited states of a hydrogen atom,respectively. According to the Bohr's model of an atom,the ratio $T_{1}: T_{2}$ is

Energy levels $A$,$B$ and $C$ of a certain atom correspond to increasing values of energy,i.e.,$E_A < E_B < E_C$. If $\lambda_1$,$\lambda_2$,and $\lambda_3$ are the wavelengths of the photons corresponding to the transitions shown in the figure,then:

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:

When a hydrogen atom transitions from $n=2$ to $n=1$, it emits a photon. The recoil speed of the atom is $\frac{x}{5} \,m/s$. Find the value of $x$. (Use: mass of hydrogen atom $= 1.6 \times 10^{-27} \,kg$)

The energy $E$ of a hydrogen atom with principal quantum number $n$ is given by $E = \frac{-13.6}{n^2} \; eV$. The energy of a photon emitted when the electron jumps from the $n = 3$ state to the $n = 2$ state of hydrogen is approximately......$eV$.

If the potential energy of a hydrogen atom in the first excited state is assumed to be zero, then the total energy of the $n = \infty$ state is,