Class 11th

The lowest value of l where 'g' orbital can be present = 4

As for any value 'n' of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1).

∴ For l = 4, the minimum of n where 'g' orbital can be present = 4+1=5.

(i) (a) 1s²  (b) 1s² 2s² 2p⁶    (c) 1s²2s²2p⁶    (d) 1s²2s²2p⁶.

(ii) (a) Na (Z = 11) has outermost electronic configuration = 3s¹

(b) N (Z = 7) has outermost electronic configuration = 2p³

(c) Cl (Z = 17) has outermost electronic configuration = 3p⁵

(iii) (a) Li  (b) P    (c) Sc

Na⁺ and Mg²⁺ are iso-electronic species having 10 electrons each. K⁺, Ca²⁺, S^2- are iso-electronic species having 18 electrons each.

Kinetic energy. K.E. = ½ mv²

=> v² = (2 x K.E.) / m

Given, K.E. = 3 x 10^-25 J = 3 x 10^-25 kg m² s^-2

Therefore, v² = [2 x (3 x 10^-25 kg m² s^-2)] / 9.1 x 10^-31 kg

=> v²= 65.9 x 10⁴ m² s^-2

=>v = 8.12 x 10² m s^-1

To calculate the wavelength of the electron

According to de Broglie's equation,

λ=h/mv  = (6.626 x 10^-34 kg m² s^-1) / (9.1 x 10^-31 kg) x (8.12 x 10² m s^-1)

= 0.08967 x 10^-5 m = 8967 x 10^-10 m = 8967Å

We know that the mass of an electron, m_e = 9.1 x 10^-31 kg,

Velocity of electron, v = 2.05 * 10⁷ m s^-1

We know that Planck's constant, h = 6.626 x 10^-34 kg m² s^-1

As per de Broglie's equation, λ=h/mv

= (6.626 x 10^-34 kg m² s^-1) / (9.1 x 10^-31 kg) x (2.05 x 10⁷ m s^-1) = 3.55 x 10^-11 m

Step I: Calculation of energy required

ΔE = E_∞– E₂

= 0 – (–2.18 * 10^-18 J) / 4 = 5.45 x 10^-19 J

Step II: Calculation of the longest wavelength of light in cm used to cause the transition

λ = hc / ΔE = (6.626 x 10^-34 J s) x (3 x 10⁸ ms^-1) / (5.45 x 10^-19 J)

= 3.644 x 10^-7 x 10² = 3.645 x 10^-5 cm.

Step I: calculation of energy required

Energy of electron (E_n) = (- 2.18 x 10^-11ergs) / n²= (- 2.18 x 10^-18 J) / n²

Energy in Bohr's 1^st orbit (E₁) = (- 2.18 x 10^-18 J) / 1²

Energy in Bohr's 5^th orbit (E₁) = (- 2.18 x 10^-18 J) / 5²

Therefore, energy required (ΔE) = E₅ – E₁ = [ (- 2.18 x 10^-18 J) / 25] – [ (- 2.18 x 10^-18 J) / 1]

 = 2.18 x 10^-18 (1 – 1/25) J

 = 2.18 x 10^-18 x 24 / 25

 = 2.09 x 10^-18 J

Step II: Calculation of wavelength of light emitted

λ = hc / ΔE = (6.626 x 10^-34 J s) x (3 x 10⁸ ms^-1) / (2.09 x 10^-18 J) = 9.50 x 10^-8 m = 950

According to Balmer formula? = 1 / λ = R_H [1/n₁² – 1/n₂²]

For the longest wavelength transition in the Balmer series of atomic hydrogen, wave number must be least. This is possible in case n₂ – n₁ = minimum; i.e. n₁ = 2 and n₂ = 3. Substituting the values:

? = 1 / λ = (1.097 x 10⁷ m^-1) [1/2² – 1/3²] = (1.097 x 10⁷ m^-1) [5/36] = 1.523 x 10⁶ m^-1

(i) For an electron, the energies in the two orbits can be compared as:

E₁ / E₂ = (n₂ / n₁)²          [Since E is inversely proportional to n²]

Given: n₁ = 1, E₁ = –2.17 * 10^-18 J atom^-1, n₂ = 5

Therefore, (–2.17 * 10^-18 J atom^-1) / E₂ = (5 / 1) ² = 25

=> E₂ = (–2.17 * 10^-18 J atom^-1) / 25 = –8.77 x 10^-20 J atom^-1

(ii) For hydrogen atom; r_n = 0.529 x n² Å

r₅ = 0.529 x (5)² = 13.225 Å = 1.3225 nm.

The maximum no. of emission lines =  [n (n–1)] /2 =  [6 (6–1)] / 2 =3 * 5 = 15

The transitions that take place are as follows: