If de Broglie wavelength of electron in the ground state of hydrogen atom is $\lambda$, what will be the de Broglie wavelength of the electron in $H{e}^{+}$ when the ion is in the second excited state?

If $r_0$ be the radius of orbit of electron in hydrogen atom in ground state, then de Broglie wavelength $\lambda =2\pi r_0$

In a hydrogen-like ion, $r_n=r_0\frac{n^2}{z}$

$\therefore$ de Broglie wavelength of electron

${\lambda }^1=\frac{2\pi r_n}{n}=2\pi r_0\frac{n}{z}=\lambda \frac{3}{2}=1.5\lambda$