According to Rydberg-Balmer equation.
\(\frac { 1 }{ \lambda } =R\left[ \frac { 1 }{ n^{ 2 }_{ 1 } } -\frac { 1 }{ n^{ 2 }_{ 2 } } \right] =R\left[ \frac { 1 }{ 1^{ 2 } } -\frac { 1 }{ n^{ 2 }_{ 2 } } \right] \)
The wavelength (\(\lambda \))will be the longest when n₂ is the smallest i.e., n₂ = 2 and 3 for two longest wavelength lines.
For n₂ = 2 : \(\frac { 1 }{ \lambda } \)= (1.097 x 10^-2 nm^-1)\(\left[ \frac { 1 }{ 1^{ 2 } } -\frac { 1 }{ 2^{ 2 } } \right] \)
= (1.097 x 10^-2 nm^-1) x 4 = 8.228 x 10^-3 nm^-1 or \(\lambda \).= 121.54 nm
For n₂ = 3 : \(\frac { 1 }{ \lambda } \)= (1 097 x 10^-2 nm^-1)\(\left[ \frac { 1 }{ 1^{ 2 } } -\frac { 1 }{ 3^{ 2 } } \right] \)
= (1.097 x 10^-2 nm^-1) x (8/9) = 9.75 x 10^-3 nm^-1 ; \(\lambda \).= 102.56 nm

The expression for the energy of electron of hydrogen is:
Eⁿ = -\(\frac { 2\pi ^{ 2 }m_{ e }^{ 4 } }{ n^{ 2 }h^{ 2 } } \)
When \(n=1,E_{ 1 }=-\frac { 2\pi ^{ 2 }m_{ e }^{ 4 } }{ (1)^{ 2 }h^{ 2 } } =-13.12\times 15^{ 5 }J \ mol^{ -1 }\)
When \(n=2,E2=-\frac { 2\pi ^{ 2 }m_{ e }^{ 4 } }{ (2)^{ 2 }h^{ 2 } } =-\frac { 13.12\times 15^{ 5 } }{ 4 } Jmol^{ -1 }\)
=- 3.28 x 10⁵ J mol^-I.
The energy required for the excitation is :
\(\Delta \)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
Kinetic energy = 1/2 mv² = 4.55 x 10^-25J = 4.55 x 10^-25 kg m² S^-2
or v² = \(\frac { 2\times KE }{ m } =\frac { 2\times (4.55\times10^{ -25 }kgm^{ 2 }s^{ -2 }) }{ (3x10^{ 6 }ms^{ -1 }) } \)=10⁶ m² S^-2
or Velocity (v) = (10⁶ m² S^-2 )^1/2 = 10³ ms^-1
Step II. Calculation of the momentum of the electron Momentum of electron = mv = (9.1 x 10^-31 kg) x (10³m s^-l) = 9.1 x 10^-28 kg m^-1
Step III. Calculation of the wavelength of the electron According to de Broglie equation:
\(\lambda \frac { h }{ mv } =\frac { (6.626\times10^{ -34 }kgm^{ 2 }s^{ -1 }) }{ (9.1x10^{ -31 }kg)\times (10^{ 3 }ms^{ -l }) } \)
= 0.728 x 10-6m = 7.28 x 10-7 m

According to de Broglie equation,\(\lambda =\frac { h }{ mv } \)
Mass of electron = 9.1 x 10^-31 kg; Planck's constant = 6.626 x 10^-34 kgm²s^-1
Velocity of electron = 1% of speed of light = 3.0 x 10⁸ x 0.01 = 3 x 10⁶ me-1
Wavelength of electron (\(\lambda \)) =\(\frac { h }{ mv } =\frac { (6.626x10^{ -34 }kgm^{ 2 }s^{ -1 }) }{ (9.1x10^{ -31 }kg)x(3x10^{ 6 }ms^{ -1 }) } \)
= 2.43 x 10^-10 m.

According to uncertainty principle,
\(\Delta x.m\Delta v=\frac { h }{ 4\pi } \) or \(m=\frac { h }{ 4\pi \Delta x\Delta v } \);h = 6.626 x 10^-34 kg m²s^-1 
\(\Delta \)x = 10^-10,m; \(\Delta \)v = 5.27 x 10^-24 ms-1
m = \(\frac { (6.626\times10^{ -34 }kgm^{ 2 }s^{ -1 }) }{ 4x3.143\times (10^{ -10 }m)\times (5.27\times10^{ -24 }ms^{ -1 }) } \)= 0.1kg

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

(i) n = 0, I = 0, \(m_{ 1 }\) = 0, \(m_{ s }\) = + 1/2
(ii) n = I, I = 0, \(m_{ 1 }\) = 0, \(m_{ s }\) = - 1/2
(iii) n = I, I = I, \(m_{ 1 }\) = 0, \(m_{ s }\) = + 1/2
(iv) n = I, 1 = 0, \(m_{ 1 }\) = + 1, \(m_{ s }\) = + 1/2
(v) n = 3 I = 3 \(m_{ 1 }\) = - 3 \(m_{ s }\) = + 1/2
(vi) n = 3 l = 1 \(m_{ 1 }\) = 0 \(m_{ s }\) = + 1/2

(i) \(_{ 17 }^{ 35 }{ X }\)
(x = Cl)    
(ii) \(_{ 92 }^{ 233 }{ X }\)
(x = U)   
(iii) \(_{ 4 }^{ 9 }{ X }\)
(x = Be)

(i)  \(_{ 17 }^{ 35 }{ CI }\)
(ii) \(_{ 92 }^{ 233 }{ U }\)
(iii) \(^{ 9 }{ Be }\)

Isoelectronic species have the same number of electrons but different atomic numbers. Number of positive charge shows the number of electrons lost and number of negative charges shows the number of electrons gained by an atom. Calculation of number of electrons have been shown below.
\(_{ 11 }Na^{ + }=11-1=10{ e }^{ - }, \ _{ 19 }K^{ + }=19-1=18{ e }^{ - },\)
\(_{ 12 }Na^{ 2+ }=12-2=10{ e }^{ - }, \ _{ 20 }Na^{ 2+ }=20-2=18{ e }^{ - },\)
\(_{ 16 }S^{ 2- }=16+2=18{ e }^{ - }, \ _{ 18 }Ar=18{ e }^{ - }\)
Hence, isoelectronic species are
\(Na^{ + }and \ Mg^{ 2+ } \ { k }^{ + },{ Ca }^{ 2+ },{ s }^{ 2- } \ and \ Ar\)