When the electron orbiting in a hydrogen atom in its ground state moves to the third excited state,the de-Broglie wavelength associated with it

The electron in a hydrogen atom is moving in an orbit of radius $0.53 \text{ Å}$. It takes $1.571 \times 10^{-16} \text{ s}$ to complete one revolution. The velocity of the electron will be $[\pi = 3.142]$.

The electron of a hydrogen atom makes a transition from the $(n + 1)^{th}$ orbit to the $n^{th}$ orbit. For large $n$,the wavelength of the emitted radiation is proportional to:

Which of the following is a property of the Rydberg constant?

Assertion $(A)$: The magnetic moment $(\mu)$ of an electron revolving around the nucleus decreases with increasing principal quantum number $(n)$.
Reason $(R)$: Magnetic moment of the revolving electron,$\mu \propto n$.

The emission series of hydrogen atom is given by $\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$ where,$R$ is the Rydberg constant. For a transition from $n_{2}$ to $n_{1}$,the relative change $\Delta \lambda / \lambda$ in the emission wavelength,if hydrogen is replaced by deuterium (assume that,the mass of proton and neutron are the same and approximately $2000$ times larger than that of electrons) is ........... $\%$