In a hydrogen atom,if the energy difference between the $n = 2$ and $n = 3$ orbits is $E$,how much energy (in terms of $E$) is required to remove an electron from the ground state?

What is the energy of an electron in a hydrogen atom in a stationary state corresponding to $n = 2$ ?

$A$ 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,

An electron in a hydrogen atom first jumps from the third excited state to the second excited state and then from the second excited state to the first excited state. The ratio of the wavelengths $\lambda_1 : \lambda_2$ emitted in the two cases is

An electron in a stationary hydrogen atom jumps from the $4^{\text{th}}$ energy level to the ground level. The velocity that the photon acquired as a result of the electron transition will be ($h=$ Planck's constant,$R=$ Rydberg's constant,$m=$ mass of the photon).

For a certain hypothetical one-electron atom,the wavelength (in $\mathring{A}$) for the spectral lines for transition from $n = p$ to $n = 1$ is given by $\lambda = \frac{1500p^2}{p^2 - 1}$ (where $p > 1$). The ionization potential of this element must be ......$V$ (Take $hc = 12420 \text{ eV-} \mathring{A}$).