The ratio of angular momentum of an electron in $\mathrm{n}^{\text {th }}$ orbit of hydrogen atom to the velocity of electron in $\mathrm{n}^{\text {th }}$ orbit is proportional to

In hydrogen atom spectrum, when an electron jumps from second excited state to the first excited state, the wavelength of radiation emitted is ' $\lambda$ '. If the electron jumps from the third excited state to the second orbit, the wavelength of radiation emitted will be $\frac{20 \lambda}{x}$. The value of $x$ is

In hydrogen spectrum, the ratio of wavelengths of the last line of Lyman series and that of the last line of Balmer series is

$$ \begin{aligned} &\text { For the following reaction, the particle ' } \mathrm{x} \text { ' is }{ }_6 \mathrm{C}^{11}\longrightarrow{ }_5 \mathrm{~B}^{11}+\beta+\mathrm{X} \end{aligned} $$