Class 11th

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:

According to Balmer formula,

Wave number (? ) = R_H [1/n₁²  - 1/n₂² ]cm^-1

= 109678 [1/2² – 1/4²] cm^-1

= (109678 x 3) / 16 cm^-1

λ = 1 /? = 16 / (109678 x 3) cm = 16 x 10⁷ / (109678 x 3) nm = 486 nm

Threshold frequency (v₀) = c /λ = (3 * 10⁸ m s^-1) / (68 x 10^-8 m) = 4.41 x 10¹⁴ s^-1

Work function (W₀) = hv₀ = (6.626 * 10^-34 Js) x (4.41 x 10¹⁴ s^-1) = 2.92 x 10^-19 J.

Energy of one photon (E) = hc / λ

= (6.626 * 10^-34 Js) x (3 * 10⁸ m s^-1) / (0.57 * 10^-6 m) = 3.48 x 10^-19 J

Rate of emission of quanta per second = Power / Energy = (25 watt) / (3.48 x 10^-19 J)

= (25 Js^-1) / (3.48 x 10^-19 J) = 7.18 x 10¹⁹ s^-1

Given: λ = 242 nm = 142 x 10^-9 m, c = 3 x 10⁸ ms^-1, h = 6.626 x 10^-34Js

We know, E = hc / λ

= (6.626 * 10^-34 Js) x (3 * 10⁸ m s^-1) / (242 * 10^-9 m) = 0.0821 x 10^-17 J

∴ Ionization energy per mol (E) = (0.0821 x 10^-17 J) x (6.022 x 10²³ mol^-1J) / 1000 = 494 kJ mol^-1