$A$ 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,

The wavelength of the radiation emitted is $\lambda_0$ when an electron jumps from the second excited state to the first excited state of a hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom,the wavelength of the radiation emitted will be $\frac{20}{x} \lambda_0$. The value of $x$ 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 total energy of an electron in the ground state of a hydrogen atom is $-13.6 \text{ eV}$. The kinetic energy of an electron in the first excited state is.....$\text{ eV}$

In a hydrogen atom,if the difference in the energy of the electron in $n = 2$ and $n = 3$ orbits is $E,$ the ionization energy of the hydrogen atom is ....... $E$.

If $13.6 \ eV$ energy is required to ionize the hydrogen atom,then the energy required to remove an electron from $n=2$ is (in $eV$)