Structure of Atoms

Greater the penetration of the electron present in a particular orbital towards the nucleus more will be the magnitude of the effective nuclear charge. Based upon this

(i) 2s orbital is closer to the nucleus than 3s orbital and hence it will experience more effective nuclear charge.

(ii) 4d orbital will experience more effective nuclear charge due to its closer proximity to 4f orbital.

(iii) 3p orbital will experience more effective nuclear charge as it is closer to the nucleus

The nuclear charge experienced by electrons depends on the distance between the nucleus and orbital. The greater is this distance, the lesser is the effective nuclear charge. Among all the p orbitals, 4p orbital lies the farthest from the nucleus and thus experiences the lowest effective nuclear charge because of the maximum magnitude of screening or shielding effect.

The electrons may be assigned to the following orbitals :
(i) 4d
(ii) 3d
(iii) 4p
(iv) 3d
(v) 3p
(vi) 4p.
The increasing order of energy is :
(v) < (ii) = (iv) < (vi) = (iii) < (i)

Since actual momentum is smaller than the uncertainty in measuring momentum, therefore, the momentum of electron cannot be defined

ν = 4.37 x 10⁵ m s^-1, m = 0.1 kg

As per de Brogile's equation,

λ= m/v = (6.626 x 10^-34 kg m² s^-1) / (0.1 kg) x (4.37 x 10⁵ m s^-1)

=6.626/0.437  x 10^-34-5 m

= 1.516 x 10^-38 m

ν = 2.19 x 10⁶ m s^-1

As per de Brogile's equation,

λ= m/v = (6.626 x 10^-34 kg m² s^-1) / (9.1 x 10^-31 kg) x (2.19 x 10⁶ m s^-1)

6.626/91 * 2.19=  x 10^-34+25 m = 0.33243 x 10^-9 m = 332.43 pm

λ = 800 pm = 800 x 10^-12 m

m = 1.675 x 10^-27 kg

As per de Brogile's equation,

ν =h/mλ  = (6.626 x 10^-34 kg m² s^-1) / (1.675 x 10^-27 kg) x (800 x 10^-12 m)

=6.626/1.675 * 8  x 10^-34+27+10

= 0.494 x 10³ ms^-1

= 494 ms^-1

As per de Brogile's equation,

λ = h / mv = (6.626 x 10^-34 kg m² s^-1) / (9.1 x 10^-31 kg) x (1.6 x 10⁶ ms^-1)

= 0.455 x 10^-9 m = 0.455 nm = 455 pm.

Radius of orbit of H like species = (0.529 / Z) n²Å = (52.9 / Z) n² pm

r₁ = 1.3225 nm = 1322.5 pm = (52.9 / Z) n₁²

r₂ = 211.6 pm = 211.6 pm = (52.9 / Z) n₂²

∴ r₁ / r₂ = 1322.5 / 211.6

=>n₁² /n₂² = 6.25

=> n₁/n₂= (6.25)^1/2 = 2.5

=> n₁ = 2.5 n₂

=> 10 n₁= 25 n₂

=> 2 n₁= 5 n₂

If n₁ = 2, then n₂ = 5. That means transition occurs from 5^th orbit to 2^nd orbit. This means that the transition belongs to Balmer series.

Now, wave number? = (1.097 x 10⁷ m^-1) x (1/2² – 1/5²) = 1.097 x 10⁷ x 21/100 m^-1 = 23.037 x 10⁵ m^-1

λ = 1/? = 1/ 23.037 x 10⁵ m^-1

= 434 x 10^-9 m = 434 nm

This transition belongs to visible region of the spectrum of light.