An electron changes its position from orbit $n = 4$ to the orbit $n = 2$ of an atom. The wavelength of the emitted radiation is ($R =$ Rydberg's constant).

In a hydrogen atom,a transition from the state $n=6$ to $n=1$ results in ultraviolet radiation. Infrared radiation will be obtained in the transition:

In terms of Rydberg constant $R,$ the shortest wavelength in the Balmer series of the Hydrogen atom spectrum will be:

In the hydrogen spectrum,the shortest and longest wavelengths of the Balmer series are $\lambda_1$ and $\lambda_2$ respectively. The Rydberg constant $R$ of hydrogen is:

Which of the following statements is true?

An electron jumps from the $4^{\text{th}}$ orbit to the $2^{\text{nd}}$ orbit of a hydrogen atom. Given Rydberg's constant $R_{H}=10^7 \ m^{-1}$,calculate the frequency in $Hz$ of the emitted radiation. (Take $c=3 \times 10^8 \ m/s$)