If the nitrogen atom had electronic configuration 1s⁷, it would have energy lower than that of the normal ground state configuration 1s² 2s² 2p³, because the electrons would be closer to the nucleus, yet 1s⁷ is not observed because it violates:

In a hydrogen atom, if energy of an electron in ground state is 13.6 eV, then that in the 2^nd excited state is

In Hydrogen atom, energy of first excited state is – 3.4 eV. Then find out KE of same orbit of Hydrogen atom.

The velocities of two particles A and B are 0.05 and 0.02 ms^–1 respectively.The mass of B is five times the mass of A. The ratio of their de-Broglie's wavelength is

The electrons, identified by quantum numbers n and l
(i) n = 4, l = 1
(ii) n = 4, l = 0
(iii) n = 3, l = 2
(iv) n = 3, l = 1
can be placed in order of increasing energy, from the lowest to highest, as:

The energy of an electron present in Bohr’s second orbit of hydrogen atom is____.

The atomic number of elements X, Y and Z are 19, 21 and 25 respectively. The number of electrons present in the 'M' shell of these elements follows the order

A body of mass 10 mg is moving with a velocity of 100 ms^-1. The wavelength of de-Broglie wave associated with it would be
(Note: h = 6.63 × 10^-34 Js)

The electronic configuration of Cr³⁺ is

If the electron of a hydrogen atom is present in the first orbit, the total energy of the electron is