The quantum numbers of four electrons are given below.
n m s
(1) Electron 1 3 0 0 -1/2
(2) Electron 2 4 0 1 1/2
(3) Electron 3 3 2 0 1/2
(4) Electron 4 3 1 0 -1/2
The correct order of decreasing energy of these electrons is -
(A) Electron 3 > Electron 1 > Electron 4> Electron 2
(B) Electron 4> Electron 2 > Electron 3 > Electron 1
(C) Electron 3 > Electron 2 > Electron 4 > Electron 1
(D) Electron 2 > Electron 4 > Electron 3 > Electron 1
(C) The correct order of decreasing energy will be electron 3 which has higher energy than 2 and in the order given.
Of the following transitions in hydrogen atom, the one which gives an absorption line of lowest frequency is
(A) n=1 to n=2
(B) n=3 to n=8
(C) n=2 to n=1
(D) n=8 to n=3
(B) Absorption line in the spectra arise when energy is absorbed i.e., electron shifts from lower to higher orbit, out of A and B, B will have the lowest frequency as this falls in the Paschen series.
The de Broglie wavelength of electron of He+ ion is 3.329Å. If the photon emitted upon de-excitation of this He+ ion is made to hit H atom in its ground state so as to liberate electron from it, what will be the de Broglie's wavelength of photoelectron ?
(A) 0.1518 Å
(B) 6.1518 Å
(C) 2.3518 Å
(D) 4.1218 Å
this is the K.E. of electron Therefore, total energy = -13.58eV
En=-13.6 x (Z2/n2)eV
For He+ion, Z=2 and En = -13.58eV
So, n2= -13.6x4 / -13.58 = n = 2.
Thus, He+ion is in the 1 excited state.
Energy of photon emitted
Energy of photon =I.P. of H + K.E.of photoelectron Thus, K.E. of photoelectron= 40.8-13.6 = 27.2eV