Class 11th

Heavy metals have a heavy nucleus and contain a large amount of positive change in their nucleus. By using heavy metals like gold and platinum in Rutherford's experiment, a large number of α-particles get deflected and experience a repulsion thus finding it hard for these α-particles to retrace their path.

If a thin foil of lighter atoms like aluminium were used in the Rutherford's experiment, the obstruction offered to the path of the fast moving α-particles would be comparatively quite less. As a result, the number of α-particles deflected will be quite less and the particles which are deflected back will be negligible.

Charge on oil droplet = – 1.282 x 10^-18C

Charge on an electron = – 1.602 x 10^-19C

Number of electrons = q /e = (– 1.282 x 10^-18C) / (– 1.602 x 10^-19C) = 8

Static electric charge (q) = 2.5 x 10^-16 C

Charge on one electron (e) = 1.602 x 10^-19 C

No. of electrons present = (2.5 x 10^-16 C) / (1.602 x 10^-19 C) = 1560

(a) The diameter of zinc atom is 2.6 Å =2.6*10⁻¹⁰m.

      The radius of Zn atom is (2.6*10⁻¹⁰) / 2=1.3*10⁻¹⁰m=130*10⁻¹²m=130 pm.

(b) The number of Zn atoms present on 1.6 cm of length are 1.6 / (2.6*10⁻⁸) =6.154*10⁷.

The length of the arrangement = 2.4 cm
Total number of carbon atoms present = 2 *10⁸

Diameter of each C-atom = (2.4 cm) / (2 x 10⁸) = 1.2 x 10^-8 cm

Radius of each C-atom = ½ x 1.2 x 10^-8 cm = 6.0 x 10^-9 cm = 0.06 nm

Length of scale = 20 cm = 20 x 10⁷ nm = 2 x 10⁸ nm

Diameter of carbon atom = 0.15 nm

∴ Number of carbon atoms which can be placed side by side in a straight line across a length of a scale of length 20 cm = (2 x 108 nm) / (0.15 nm) = 1.33 x 10⁹

The expression for the ionization energy atom,

E_n = (2.18 * 10^-18 x Z²) / n² J atom^-1

For H atom, (Z = 1). So,

E_n =2.18 * 10^-18 * (l)² J atom^-1  (given)

For He⁺ ion (Z = 2). So,

E_n =2.18 * 10^-18 * (2)² = 8.72 * 10^-18 J atom^-1  (one electron species)

For an atom? = 1/ λ = R_H Z² [ (1/n₁²) – (1/n₂²)]

For He⁺ spectrum: Z = 4, n₂ = 4, n₁ = 2.

∴? = 1/ λ = R_H Z² [ (1/2²) – (1/4²)]

= R_H 2² [ (1/2²) – (1/4²)]

= 3R_H /4

For hydrogen spectrum:

∴? = 1/ λ = R_H 1² [ (1/n₁²) – (1/n₂²)] = 3R_H /4

=> (1/n₁²) – (1/n₂²) = 3/4

This corresponds to n₁ = 1 and n₂ =2 and means that the transition has taken place in the Lyman series from n = 2 to n =1.

According to Bohr's theory,

mvr = nh / 2π

=> 2πr = nh/mv

=> mv = nh / 2πr - (i)

According to de Broglie equation,

 h / λ - (ii)
Comparing equations (i) and (ii)

nh / 2πr = h / λ

=> 2πr = n λ

Thus, the circumference (2πr) of the Bohr orbit for hydrogen atom is an integral multiple of the de Broglie wavelength.

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