In a hydrogen atom,the electron is in the $n^{th}$ excited state. It may come down to the second excited state by emitting ten different wavelengths. What is the value of $n$?

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

$\frac{x}{x+4}$ is the ratio of energies of photons produced due to the transition of an electron of a hydrogen atom from its $(i)$ third permitted energy level to the second level and $(ii)$ the highest permitted energy level to the second permitted level. The value of $x$ will be.

As per the given figure,$A$,$B$,and $C$ are the first,second,and third excited energy levels of a hydrogen atom,respectively. If the ratio of the two wavelengths (i.e.,$\frac{\lambda_1}{\lambda_2}$) is $\frac{7}{4n}$,then the value of $n$ will be

When a hydrogen atom emits a photon during a transition from $n=4$ to $n=2$,its recoil speed is about (in $m \ s^{-1}$)

The ratio of energies of photons produced due to transition of an electron in a hydrogen atom from the second energy level to the first energy level and from the fifth energy level to the second energy level is: