Tips and tricks to solve the nunrical problems of structure of atom
Answers
Answered by
3
All lines in the visible region are of Balmer series but reverse is not true i.e., all Balmer lines will not fall in visible region.
* A part of an atom up to penultimate shell is a kernel or atomic core.
* If the energy supplied to hydrogen atom is less than 13.6 eV it will awpt or absorb only those quanta which can take it to a certain higher energy level i.e., all those photons having energy less than or more than a particular energy level will not be absorbed by hydrogen atom, but if energy supplied to hydrogen atom is more than 13.6eV then all photons are absorbed and excess energy appear as kinetic energy of emitted photo electron.
* No of nodes in any orbital = (n-l-1)
* No of nodal planes in an orbitals = l
* The d orbital which does not have four lobes is dz2
* The d orbital whose lobes lie along the axis is dx2-y2
* Spin angular momentum = √s(s+1)(h/2π)
* Total spin = + (n/2); where n is no of unpaired e-
* Magnetic moment = √n(n+1) B.M. (Bohr magnetron) of n unpaired e-
* Ion with unpaired electron in d or f orbital will be coloured.
* Exception of E.C. are Cr(24), Cu(29), Mo(42), Ag,(47),W(74), Au(79)
* No. of waves n = (2πr)/λ (where λ=h/mv)
* No. of revolutions of e- per second is v/2πr
* The solution of schrodinger wave equation gives principal, azimuthal and magnetic quantum numbers but not the spin quantum number.
* A part of an atom up to penultimate shell is a kernel or atomic core.
* If the energy supplied to hydrogen atom is less than 13.6 eV it will awpt or absorb only those quanta which can take it to a certain higher energy level i.e., all those photons having energy less than or more than a particular energy level will not be absorbed by hydrogen atom, but if energy supplied to hydrogen atom is more than 13.6eV then all photons are absorbed and excess energy appear as kinetic energy of emitted photo electron.
* No of nodes in any orbital = (n-l-1)
* No of nodal planes in an orbitals = l
* The d orbital which does not have four lobes is dz2
* The d orbital whose lobes lie along the axis is dx2-y2
* Spin angular momentum = √s(s+1)(h/2π)
* Total spin = + (n/2); where n is no of unpaired e-
* Magnetic moment = √n(n+1) B.M. (Bohr magnetron) of n unpaired e-
* Ion with unpaired electron in d or f orbital will be coloured.
* Exception of E.C. are Cr(24), Cu(29), Mo(42), Ag,(47),W(74), Au(79)
* No. of waves n = (2πr)/λ (where λ=h/mv)
* No. of revolutions of e- per second is v/2πr
* The solution of schrodinger wave equation gives principal, azimuthal and magnetic quantum numbers but not the spin quantum number.
Similar questions