The ratio of energies of photons produced due to the transition of an electron in a hydrogen atom from its $(i)$ second to first energy level and (ii) highest energy level to second energy level is: (in $3:1$)

As the $n$ (number of orbit) increases,the difference of energy between the consecutive energy levels

Three energy levels of a hydrogen atom and the corresponding wavelengths of the emitted radiation due to different electron transitions are as shown. Then,

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

The maximum wavelength of incident radiation required to ionize a hydrogen atom in its ground state is nearly

$A$ 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,