Schrodinger Equation
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Study the extract of the periodic table given below and answer the questions that follow. Give the alphabet corresponding to the elements in the question. DO NOT repeat an element .
A | |||||||||||||||||
C | D | E | |||||||||||||||
B | G | F | |||||||||||||||
- Which element forms electrovalent compound with G?
- The ion of which element will migrate towards the cathode during electrolysis?
- Which non metallic element has the valency of 2?
- Which is an inert gas?
- [Ni(NO2)6]4−<[Ni(NH3)6]2+<[Ni(H2O)6]2+
- [Ni(NO2)6]4−<[Ni(H2O)6]2+<[Ni(NH3)6]2+
- [Ni(H2O)6]4−<[Ni(NH3)6]2+<[Ni(NO2)6]4−
- [Ni(NH3)6]2+<[Ni(H2O)6]2+<[Ni(NO2)6]4−
What is the symbol of aluminium ion?
(a0 is first Bohr radius)
Which of the following option(s) is/are correct?
- Orbital may be 3p
- Angular node is at 2a0 from nucleus
- Ratio of density of probability of finding electron at distance a0 to nucleus is e−14
- Orbital has a total of one node
(a0 is first Bohr radius)
Ψ=√2r81a20√πa0(6−ra0)e−r3a0 cosθ
List-I List-II(I)Which orbital is this?(P)1(II)Number of total node(s)(Q)2(III)Nodal plane(R)3pz(IV)Number of radial node(s)(S)3px(T)XY plane(U)YZ plane
Match the correct combination considering List-I and List-II
- (III), (T) and (IV), (P)
- (III), (U) and (IV), (P)
- (III), (T) and (IV), (Q)
- (III), (U) and (IV), (Q)
- Wave functions are found by solving Schrodinger wave equation
- Two spectral lines of an element must have the same wave number
- Energy of an electron at infinite distance is zero yet it is maximum
- The position and momentum of a large rolling ball can be measured accurately.
[Co(H2O)6]3+;[Co(CN)6]3−;[Co(I)6]3−;[Co(en)3]3+IIIIIIIV
- III > I > IV >II
- II > IV > I >III
- III > I > II >IV
- I > III > IV >II
In a H - atom, the transition takes place form L to K shell. If R = 1.08×107m−1. The wave length of the light obsorbed in nearly:
1850 A0
1650 A0
4400 A0
1250 A0
(a0 is first Bohr radius)
Ψ=√2r81a20√πa0(6−ra0)e−r3a0 cosθ
List-I List-II(I)Which orbital is this?(P)1(II)Number of total node(s)(Q)2(III)Nodal plane(R)3pz(IV)Number of radial node(s)(S)3px(T)XY plane(U)YZ plane
Match the correct combination considering List-I and List-II
- (I), (R) and (II), (P)
- (I), (S) and (II), (Q)
- (I), (S) and (II), (P)
- (I), (R) and (II), (Q)
- 4
- 3
- 2
[Co(H2O)6]3+;Co(CN)6]3−;[CoI6]3−;[Co(en)3]3+ I II III IV
- III>I>IV>II
- II>IV>I>II
- III>I>II>IV
- I>III>IV>II
A permissible solution to the schrodigner wave equation gave an idea of _____ quantum numbers.
4
1
2
3
The wave function ψ in the Schrodinger wave equation represents
Probability of the electron
Amplitude of the wave
Frequency of the wave
Speed of the wave
What is the formula for a sine wave?
- Spherical node appears
- Electron probability density is highest in the vicinity of the nucleus
- Probability density drops to zero after maximum probability is reached
- Probability density rises to second highest value
Ψ1s=π√2e−r/a0
where a0= Radius of first Bohr orbit
r= Distance from the nucleus (Probability of finding the electron varies with respect to it)
- e
- e2
- 1/e2
- Zero
Quantum numbers are
Solution to schrodinger equation
Variable in the atomic model
Coordinates of Electron position in an atom
State variables of electrons
The possibility of finding an electron in an orbital was conceived by
Rutherford
Bohr
Heisenberg
Schrodinger
Name the following :
Formula of the compound formed between Aluminum and Oxygen
- 2s orbital
- 3s orbital
- 1s orbital
- 2p orbital
What was Schrodingers discovery?
ψ1s=π√2e−r/a0
where, a0= Radius of first Bohr orbit
r= Distance from the nucleus (Probability of finding the electron varies with respect to it)
What will be the ratio of probabilities of finding the electrons at the nucleus to first Bohr's orbit a0?
- e2
- e
- 1e2
- Zero
ψ=k(1ao)3/2[(x−1)(x2−8x+12)]e−x/2
x=2rao, ao is radius of the first Bohr orbit.
- orbitals
- wave function
- amplitude function
- both B & C