The de Broglie wavelength of the electron in the ground state is $\lambda_1$ and that in the $n = 3$ level is $\lambda_3$. Then $\lambda_3$ is given by:

The third line of the Balmer series of an ion equivalent to a hydrogen atom has a wavelength of $108.5 \ nm$. The ground state energy of an electron of this ion will be.....$eV$

In the lowest energy level of a hydrogen atom,the electron has an angular momentum of:

In the hydrogen atom, the electron makes a transition from the higher orbit $(i)$ to a lower orbit $(f)$. The ratio of the radius of the orbits is given by $r_i : r_f = 16 : 4$. The wavelength of the photon emitted due to this transition is . . . . . . nm. (Given Rydberg constant $R = 1.0973 \times 10^7 \text{ m}^{-1}$)

If the energy of a hydrogen atom in the $n^{th}$ orbit is ${E_n}$,then the energy in the $n^{th}$ orbit of a singly ionized helium atom will be:

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 ........... $\%$