Given  \(\Delta\)x = 0.1  or 0.1 x 10^-10 m or 10^-11 m.
h = 6.626 x 10^-34 kg m² s^-1, m = 9.11 x 10^-31 kg
According to uncertainty principle,
\(\Delta x,m(\Delta v)=\frac{h}{4\pi}\)
i.e., \(\Delta v=\frac{h}{4\pi\times m\times \Delta x}\)
= \(\frac{6.626\times 10^{-34}kg m^2 s^{-1}}{4\times 3.14\times 9.11\times 10^{-31}kg \times 10^{-11}m}\)
= 5.79 x 10⁶ ms^-1.

KE = \(\frac{1}{2}\)mv² = 5.85 x 10^-25 J
By de Broglie equation \(\lambda=\frac{h}{mv}\)
But \(\lambda=\frac{\nu}{V}=\frac{h}{mv}\)
v = \(\frac{mv^2}{h}=\frac{2\times 5.85\times 10^{-25}J}{6.026\times 10^{-34}J}\)
= 1.77 x 10⁹ s^-1.

Given m = 9.1 x 10^-31 kg; v = 10³ ms^-1
h = 6.6 x 10^-34 kg m² sec^-1
\(\lambda=\frac{h}{mv}=\frac{6.6\times 10^{-34}kgm^2 sec^{-1}}{9.1\times 10^{-31}kg\times 10^3 ms^{-1}}\)
= 7.25 x 10^-7 m.

The reasons are:
(i) Symmetrical distribution: 3d⁵(half-filled) and 3d¹⁰ (completely filled) one more symmetrical and hence more stable.
(ii) Exchange energy: Electrons with parallel spins in degenerate orbitals tend to exchange their positions. As a result, energy is released. This energy is called exchange energy. Greater the exchange energy, greater is the stability.

Angular momentum of electron in any orbital = \(\sqrt{l(l+1)}\times\frac{h}{2\pi}\)
(i) for 2s orbital, l = 0
Therefore angular momentum = \(\sqrt{0(0+1)}\times\frac{h}{2\pi}=0\)
(ii) for 4f orbital, I = 3
Therefore angular momentum = \(\sqrt{3(3+1)}\times\frac{h}{2\pi}\)
= \(2\sqrt{3}\frac{h}{2\pi}\)
\(=\sqrt{3}\frac{h}{\pi}\)

(i) In AI, 2p should be filled before filling 3s, orbital states.
\(\therefore\) correct electronic configuration is 1s² 2s² 2p⁶ 3s² 2p¹
(ii) In B, total electrons = 5. Electronic configuration is 1s² 2s² 2p¹
(iii) In F, total electrons = 9. Electronic configuration is 1s² 2s² 2p⁵.

The electronic configuration of sulphur is 1s² 2s² 2p⁶ 3s² 3p⁴
For 1s², n,+ l = 1+ 0 = 1 ; For 2p⁶, n + 1 = 2 + 1 = 3
2s², n + I = 2 + 0 = 2;      For 3s², n + I = 3 + 0 = 3
3p⁴, n + 1 = 3 + 1 = 4
Thus n + 1 = 3 for 2p⁶ electron and 3s² electron = 8 electrons.

(i) 2s > 1s
(ii) 3p > 2p
(iii) 3d_xy > 3d_yz
(iv) 3s = 3d
(v) 5s > 4f.

No. of radial nodes = n - I +1 = 5 - 3 - 1 = 1.

Similarities:
(i) Both have similar shape.
(ii) Both have same angular momentum = \(\sqrt{l(l+1)} \frac{h}{2\pi}\)
Dissimilarities:
(i) Is orbital has no node while 2s orbital has one node.
(ii) Energy of 2s orbital is greater than Is orbital.
(iii) The size of the 2s orbital is larger than Is orbital.

0.529 \(\overset { o }{ A } \), Bohr radius.

This is because, according to this principle, if one electron of an atom has same particular values, for the four quantum numbers, then all the other electrons in that atom are excluded from having the same set of values.

Subshells with n + 1 = 5 are 5s, 4p, 3d. Hence electrons present are 2 + 6 + 10 = 18.

(i) The last electron is the 'one' electron present in 'p' sub level.
i.e., n = 2; l = 1; m = -1,0 or + 1,s = + 1/2 or - 1/2
(ii) If the first electron present in 'p' subshell has the-following quantum numbers.
n = 3, l = 1, m = -1,0 (or) + 1, s = + 1/2,
the last electron has n = 3, l = 1, m = -1, 0 (or) + 1, s = - 1/2.

Hydrogen atom (H): Electronic configuration = IS1.
The four quantum number of electrons present in a hydrogen atom is
n = 1 l = 0; m = 0; s = +\(\frac{1}{2}\)
The four quantum number of two electrons present in a helium atom is
n = 1; l = 0; m = 0; s = + \(\frac{1}{2}\)
n = 1; l = 0; m = 0; s = + \(\frac{1}{2}\)

H₂ = 1 + 1 = 2e^-
H⁺₂; = 2 - 1 = le^-
\({O}_{2}\) = 8 + 8 = 16e^-
\({O}_{2}^{-}\) = 8 + 8 + 1= 17e^-

(i) Mass number is a whole number because it is the sum of number of protons and number of neutrons.
(ii) Atomic mass is fractional because it is the average relative mass of its atom as compared with mass of an atom of C-12 isotope taken as 12.

The charge of an electron is 1.602 x 10^-19 coulomb
The mass of an electron is 9.11 x 10-³¹ kg

According to J. C. Maxwell, whenever an electron is subjected to acceleration, it emits radiation and loses energy. As a result of this, its orbit should become smaller and smaller and finally it should drop into the nucleus by following a spiral path. This means that atom would collapse and thus Rutherford's model failed to explain stability of atoms. Another drawback of the Rutherford's model is that it gives no information about the electronic structure of an atom.

(i) Atoms consist of huge positively charged centers called nuclei.
(ii) Most of the space inside the atom is empty.

According to this theory, atom was assumed to consist of a sphere of uniform distribution of about 10^-10 m positive charge with electrons embedded in it such that the number of electrons equal to the number of positive charges and the atom as a whole is electrically neutral.

\(\mathrm{E}_{n}=\frac{-13.6}{\mathrm{n}^{2}} \mathrm{eV}\)
Second excited state
\(\therefore E_{3}=\frac{-13.6}{9} \mathrm{eV}\)
n = 3
E₃ = -1.51 eV