Hydrogen has three isotopes (A), (B) and (C). If the number of neutron(s) in (A), (B) and (C) respectively, are (x), (y) and (z), the sum of (x), (y) an (z) is :

For the Balmer series in the spectrum of H atom,
$$\overline \nu = {R_H}\left\{ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right\}$$, the correct statements among (I) to (IV) are :
(I) As wavelength decreases, the lines in the series converge
(II) The integer n₁ is equal to 2
(III) The lines of longest wavelength corresponds to n₂ = 3
(IV) The ionization energy of hydrogen can be calculated from wave number of these lines

The number of orbitals associated with quantum number n = 5, m_s = +$${1 \over 2}$$ is :

The electrons are more likely to be found :