Electronic configuration of Fe³⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰ 3d⁵
Electronic configuration of Ni²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰ 3d⁸
Fe³⁺ has stable 3d⁵ half filled configuration.

(i) In a multi-electron atom, in addition to the electrostatic attractive forces between the electron and nucleus, there exists a repulsive force among the electrons.
(ii) These two forces are operating in the opposite direction. This results in the decrease in the nuclear force of attraction on electron.
(iii) The net charge experienced by the electron is called effective nuclear charge.
(iv) The effective nuclear charge depends on the shape of the orbitals and it decreases with increases in azimuthal quantum number l.
(v) The order of the effective nuclear charge felt by a electron in an orbital within the given shell is s > p > d > f.
(vi) Greater the effective nuclear charge, greater is the stability of the orbital. Hence, within a given energy level, the energy of the orbitals are in the following order s < p < d < f.

Step 1. Calculation of energy required
The energy required is the difference in the energy when the electron jumps from orbit with n = \(\infty\) to orbit with n = 2.
The energy required (\(\triangle\)E) = E_\(\infty\)-E₂
\(=0-(-{2.18\times 10^{-18}\over 4}J)=5.45\times 10^{-19}J\)
Step II. Calculation of the longest wavelength of light in cm used to cause the transition
\(\triangle E=hv=hc/ \lambda\)
\(\lambda={hc\over \triangle E}={(6.626\times 10^{-34}J s)\times (3\times 10^{8}ms^{-1})\over (5.45\times 10^{-19}J)}\)
= 3.644 x 10^-7 m = 3.644 x 10^-7 x 10² = 3.645 x 10^-5 cm.

An element can be identified by its atomic number only. Let us find the atomic number.
Let the number of protons = x
\(\therefore\) Number of neutrons = \(x+{x\times 31.7\over 100}=(x+0.317x)\)
Now, Mass no. of element = 35, No. of protons + No. of neutrons
81 = x + x + 0.317x = 2.317x = 2.317x or x = \({81\over 2.317}=35\)
\(\therefore\) No. of protons = 35, No. of neutrons = 81 - 35 = 46
Atomic number of element (Z) = Number of protons = 35
The element with atomic number (Z) 35 is bromine \(^{81}_{35}Br.\)

\(\lambda ={h\over mv}; \lambda ={(6.626\times 10^{-34}Js)\over (9.1\times 10^{-31}kg)\times (1.6\times 10^{6}ms^{-1})}\)
= 0.455 x 10^{-34+ 25}m = 0.455 nm = 455 pm.

The set of quantum number is possible.

The set of quantum number is not possible because, for n = 3,\(l\ne3.\)

The set of quantum numbers is not possible because for l = 0,m_l cannot be +1. It must be zero.

The set of quantum numbers is not possible because, for n = 1,1 cannot be equal to 1. It can have 0 value.

The set of quantum numbers is possible.

The set of quantum numbers is not possible because the minimum value of n can be 1 and not zero.

\(E_n=-{2\pi^{2}m_e^4\over n^2h^2}\)
When n = 1,E₁\(=-{2\pi^{2}m_e^4\over (1)^2h^2}=-13.12\times 10^5 J\ mol^{-1}\)
When n = 2,E₂ \(=-{2\pi^{2}m_e^4\over (2)^2h^2}={-13.12\times 10^5 \over 4}J\ mol^{-1}\)
= 3.28 x 10⁵ J mol^-1.
The energy required for the excitation is:
\(\triangle\)E = E₂ - E₁ =(-3.28 x 10⁵) - (- 13.12 x 10⁵) = 9.84 x 10⁵ J mol^-1.

Step I. Calculation of the velocity of electron
 Energy (kinetic energy) of electron = 1.0 x 10⁴ volts.
= 1.0 x 10⁴ x 1.6 x 10^-19 J = 1.6 x 10^-15J
= 1.6 x 10^-15kg m² s^-2
or 1\2 mv² = 1.6 x 10^-15kg m² S^-2 
or v= \(({2\times 1.6\times 10^{-15}kg m^2s^{-2}\over 9.1\times 10^{-31}kg})^{1\over2}=5.93\times10^{7}ms^{-1}.\)
Step II. Calculation of wavelength of the electron
According to de Broglie equation,
\(\lambda ={h\over mv};\lambda={(6.626\times 10^{-34}kg m^{2}s^{-1})\over (9.1\times 10^{-31}kg)\times(5.93\times10^{7}ms^{-1})}=1.22\times 10^{-11}m\)

According to de Broglie equation, \(\lambda ={h\over mv}\)
Mass of electron = 9.1 x 10^-31kg; Planck's constant = 6.626 x 10^-34kg m² s^-1
Velocity of electron = 1% of speed of light = 3.0 x 10⁸ x 0.01 = 3 x 10⁶ ms^-1
Wavelength of electron(\(\lambda\)) =\(h\over mv\) = \({(6.626\times 10^{-34}kg m^2s^{-1})\over (9.1\times 10^{-31}kg)\times(3\times 10^6ms^{-1})}\)
= 2.43 x 10^-10m.

(i) Boron (Z = 5) ; 1s² 2s² 2p¹
(ii) Neon (Z = 10) ; 1s² 2S²2p⁶
(iii) Aluminium (Z = 13) ;1s² 2S²2p⁶ 3s² 3p¹
(iv) Chlorine(Z = 17) ; 1s² 2s² 2p⁶ 3s² 3p⁵
(v) Calcium (Z = 20) ;1s² 2S²2p⁶ 3s² 3p⁶ 4s²
(vi) Rubidium (Z = 37) ; 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 5s¹

According to de Broglie equation, \(\lambda ={h\over mv}\)
Momentum of electron, mv =\({h\over \lambda}={({6.626\times 10^{-34}}kg m^2s^{-1})\over (5200\times 10^{-10}m)}\)
=1.274 x 10^-27kg ms^-1.... (i)
The momentum of electron can also be calculated as = mv = (9.1 x 10^-31kg) x v ....(ii)
Comparing (i) and (ii)
(9.1 x 10^-31kg) x v = (1.274 x 10^-27kg ms^-1)
\(v={(1.274\times 10^{-27}kg \ ms^{-1})\over (9.1 \times 10^{-31}kg) }=1.4\times 10^{3}ms^{-1}\)

According to uncertainty principle,
\(\triangle x.m\triangle v={ h4\pi }\) or \(\triangle m={ h4\pi \triangle x\triangle v };h=6.626\times 10^{ -34 }kg\ m^{ 2 }s^{ -1 }\)
\(\triangle x=10^{-10}m;\triangle x=5.27\times 10^{-24}ms^{-1}\)
\(m={(6.626\times 10^{-34}kgm^2s^{-1} )\over 4\times 3.143\times (10^{-10}m)\times (5.27\times 10^{-24})ms^{-1}}=0.1\ kg\)

Step I. Calculate of the velocity of electron
Kinetic energy \(=\frac{1}{2}\ mv^2=3.0\times10^{-25}J=3.0\times10^{-25}\ Kg\ m^2s^{-2}\)
\(v^2=\frac{2\times K.E.}{m}=\frac{2\times(3.0\times10^{-25}\ Kg\ m^2s^{-2})}{(9.1\times10^{-31}\ Kg)}=65.9\times 10^4\ m^2s^{-2}\)
\(v=(65.9\times10^4ms^{-2})^{\frac{1}{2}}=8.12\times10^2 \ ms^{-1}\)
Step II. Calculation of wavelength of the electron According to de Broglie's equation,
\(\lambda=\frac{h}{mv}=\frac{(6.626\times 10^{-34}\ Kg\ m^2s^{-1})}{(9.1\times10^{-31\ Kg})\times(8.12\times10^2\ ms^{-1})}\)
= 0.08967\(\times\)10^-5m = 8967\(\times\)10^-10m = 8967\(\mathring A\)\((\therefore 1\mathring A=10^{-10}m)\)

According to de Broglie's equation,\(\lambda=\frac{h}{mv}\)
Mass of electron (m) = 9.1 x 10^-31 kg
Velocity of electron (v) = 2.05 x 10⁷ ms^-1
Planck's constant (h) = 6.626 x 10^-34 kg m²s^-1
\(\lambda=\frac{(6.626\times10^{-34}\ Kg\ m^2s^{-1})}{(9.1\times10^{-31}\ Kg)\times(2.05\times10^7\ ms^{-1})}=3.55\times10^{-4}\ m.\)

(i) It is represented by the letter 'l' and can take integral values from zero to n - 1, where n is the principal quantum number.
(ii) Each l value represents a subshell (orbital). l = 0, 1, 2, 3 and 4 represents the s, p, d, f and g orbitals respectively.
(iii) The maximum number of electrons that can be accommodated in a given subshell (orbital) is 2(2l + 1).
(iv) It is used to calculate the orbital angular momentum using the expression Angular momentum \(=\sqrt {l(l+1)}\frac{h}{2\pi}\)

(i) The principal quantum number represents the energy level in which electron revolves around the nucleus and is denoted by the symbol 'n'.
(ii) The 'n' can have the values 1,2,3, ... n = 1 represents K shell; n = 2 represents L shell and n = 3, 4, 5 represent the M, N, O shells, respectively.
(iii) The maximum number of electrons that can be accommodated in a given shell is 2n².
(iv) 'n' gives the energy of the electron,
\(E_n=\frac{(-1312.8)Z^2}{n^2}\ kJ\ mol^{-1}\) and the distance of the electron from the nucleus is given by \(r_n=\frac{(0.529)n^2}{Z}A\)

Uncertainty in position \(=\triangle x\)
\(=\frac{0.5\%}{100\%}\times0.529\mathring A\)
\(=\frac{0.5}{100}\times10^{-10}\times0.529\ m\)
\(\triangle x=2.645\times10^{-13}\ m\)
From Heisenberg's uncertainty principle
\(\triangle x.\triangle p\ge\frac{h}{4\pi}\)
\(\triangle x.m.\triangle v \ge\frac{h}{4\pi}\)
\(\triangle v\ge\frac{h}{\triangle x.m.4\pi}\)
\( \geq \frac{6.626 \times 10^{-34}}{4 \times 3.14} \times \frac{1}{2.645 \times 10^{-13} \times 9.11 \times 10^{-31}} \)
\(\triangle v=2.189\times10^8\ ms^{-1}\)

(i) The wave nature of electron-was experimentally confirmed by Davisson and Germer.
(ii) They allowed the accelerated beam of electrons to fall on a nickel crystal and recorded the diffraction pattern.
(iii) The resultant diffraction pattern is similar to the X-ray diffraction pattern.
(iv) The finding of wave nature of electron leads to the development of various experimental techniques such as electron microscope, low energy electron diffraction etc.

(i) \(\lambda_{iron\ ball}=\frac{h}{mv}\)
\(=\frac{6.626\times10^{-34}\ Kg\ m^2s^{-1}}{6.626\ Kg\times10ms^{-1}}\)
\(=1\times10^{-35}\ m\)
(ii) \(\lambda_{electron}=\frac{h}{mv}\)
\(=\frac{6.626\times10^{-34}\ Kgm^2s^{-1}}{9.11\times10^{-31}\ Kg\times72.73\ ms^{-1}}\)
\(=\frac{6.626}{6.626}\times10^{-3}\ m=1\times10^5\ m\)
For an electron, the de Broglie wavelength is significant and measurable while for iron ball it is too small to measure, hence it becomes insignificant.

(i) The Bohr's atom model is applicable only to species having one electron such as hydrogen, Li²⁺ etc and not applicable to multi-electron atoms.
(ii) It was unable to explain the splitting of spectral lines in the presence of magnetic field (Zeeman effect) or an electric-field (Stark effect).
(iii) Bohr's theory was unable to explain why the electron is restricted to revolve around the nucleus in a fixed orbit in which the angular momentum of the electron is equal to \(\frac{nh}{2\pi}\) 

(i) Rutherford bombarded a thin gold foil with a stream of fast moving a-particles.
(ii) It was observed that most of the \(\alpha\)-particles passed through the foil.
(iii) Some of them were deflected through a small angle.
(iv) Very few α-particles were reflected back by 180°.
(v) Based on these observations, he proposed that in an atom, there is a tiny positively charged nucleus and the electrons are moving around the nucleus with high speed.

According to uncertainty principle,
\(\triangle x.m\triangle v={h\over 4\pi}\) or \(\triangle v={h\over 4\pi m \triangle x};\) h = 6.626 x 10^-34 kg m² s^-1;m = 10g = 10^-2 kg
\(\triangle x =10^{-5}m;\triangle v ={(6.626\times 10^{-.34}kgm^2s^{-1})\over4\times 3.143\times (10^{-2}kg)\times (10^{-5}m)}=5.27\times 10^{-28}\)mv

Let the no. of electrons in the ion = x
\(\therefore\) the no. of the protons = x + 3 (as the ion has three units positive charge) and the no. of neutrons = \(x+{30.4x\over 100}=x+0.304x\)
Now, mass number of ion = Number of protons + Number of neutrons
= (x + 3) + (x + 0.304 x)
\(\therefore\) 56 = (x + 3) + (x + 0.304 x) or 2.304 x = 56 - 3 = 53
\(x={53\over 2.304}=23\)
Atomic number of the ion (or element) = 23 + 3 = 26
The element with atomic number 26 is iron (Fe) and the corresponding ion is Fe³⁺.

According to Bohr's theory,
\(mvr={nh\over 2\pi}\)       (n = 1, 2, 3, ... so on)
or \(2\pi r={nh\over mv}\)or \( mv={nh\over 2\pi r}\) ......(i)
According to de Broglie equation,
\(\lambda ={h\over mv}\)or \(mv ={h\over \lambda}\).......(ii)
Comparing (i) and (ii),
\({nh\over 2\pi r}={h\over \lambda}\) or ​​​​​​​\( 2\pi r=n\lambda\)
Thus, the circumference (\(2\pi\)r) of the Bohr orbit for hydrogen atom is an into the de Broglie wavelength.

(a) For n = 4
Total number of electrons= 2n² = 2 x 16 = 32
Half out of these will have m_s = - 1/2
\(\therefore\) Total electrons with m_s (- 1/2) = 16
(b) For n=3
1= 0 ; m_l = 0, m = +1/2 - 1/2 (two e^-)