Structure of Atoms

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^–)

(a) The set of quantum numbers is not possible because the minimum value of n can be 1 and not zero.
(b) The set of quantum numbers is possible.
(c) The set of quantum numbers is not possible because, for n = 1, l cannot be equal to 1. It can have 0 value.
(d) The set of quantum numbers is possible.
(e) The set of quantum numbers is not possible because, for n = 3, l cannot be 3.
(f) The set of quantum numbers is possible

(a) 1s orbital

(b) 4f orbital

(c) 3p orbital

(d) 4d orbital

(i) For n = 3; l = 0, 1 and 2.

For l = 0 ; m_l = 0

For l = 1; m_l = +1, 0, -1

For l = 2 ; m_l = +2, +1,0, +1, + 2

(ii) For an electron in 3rd orbital ; n = 3; l = 2 ; m_l can have any of the values -2, -1, 0, + 1, +2.

(iii) For 1p orbital, n and l are both equal to 1. Since, l should always have a lower value than n. So, 1p ortial does not exist.

For 3f orbital, n=3 and l=3. For the same reason, the existence of 3f orbital is not possible. 1p and 3f orbitals are not possible

Number of electrons in:

(i) H₂⁺ = 1

(ii) H₂ = 2

(iii) O₂⁺ = 15