Physics - Modern Physics Question with Solution | TestHub

We know that wavelength is inversely proportional to momentum, let ${\lambda }_B$ is de-Broglie Wavelength corresponds to momentum $P_B$, similarly  let ${\lambda }_G$ is Wavelength at ground state corresponds to momentum$P_G$, 
$\frac{{\lambda }_B}{{\lambda }_G}= \frac{P_G}{P_B}= \frac{m{v}_G}{m{v}_B}$, and velocity, $V\propto \frac{z}{n}$, where $Z$ is the atomic number, and $n$ is a number of orbits, here, $n_G=1$ (ground state), $n_B=3$ (second excited state). So, $\frac{{\lambda }_B}{{\lambda }_G}= \frac{n_B}{n_G}= \frac{3}{1}$$\Rightarrow {\lambda }_B=3{\lambda }_G$