Solved Examples: Structure of Atom | Chemistry for JEE Main & Advanced PDF Download

Isotopes contain the same number of protons.

The number of angular node is equal to the value of l (azimuthal quantum number).

The value of m cannot be greater than the value of l.

Orbitals having the same value of n + l will have the same energy irrespective of the value of m.

The outer filled orbital in ₄₇Ag is 4d. Its electronic configuration is [Kr](4d)¹⁰(5s)1.

The electronic configuration of ₂₈Ni²⁺ (electron = 26) is (1s)²(2s)²(2p)⁶(3s)²(3p)⁶(3d)⁸.

The energy of a single electron species such as H, He⁺, and Li²⁺ depends only on the principal quantum number whereas for a multielectron species, it depends on the principal as well as azimuthal quantum number.

It is ψ² and not ψ is proportional to the probability of finding an electron.

Binding energy of electron per atom of the metal

The expected outer electronic configuration of ₅₈Ce is (4f)¹(5d)¹(6s)². But due to very similar energies of 4f and 5d orbitals in all the lanthanides excepting Gd (4f⁷5d¹6s²) and ₇₁Lu (4f¹⁴, 5d¹, 6s²), (5d)1 electron is shifted to (4f) orbital and hence 5d orbital remains unoccupied. Thus, the observed configuration of ₅₈Ce is (4f)²(5d)⁰(6s)².

The expression of radius is

where K  is constant.

First line in the Lyman series of hydrogen atom corresponds to

For He⁺, the spectral line occurs at

IE = N_AR_∞ hc = (6.022 x 10²³ mol^–1) (1.09678 x 10⁷ m^–1) (6.626 x 10^–34 J s) (3 x 10⁸ m s^–1)  
= 1.313 x 10⁶ J mol^–1 = 1313 kJ mol^–1

Paschen spectral series lies in the near infrared region of electromagnetic radiation.

E_Li = Z²EH = (3²) (2.17 x 10^–11 erg) = 1.953 x 10^–10 erg = 1.953 x 10^–17 J

We have
63.546 u = x (62.930 u) + (1 – x) (64.928 u)
Hence

Mass percent of ⁶⁵Cu = 0.31 x 100 = 31%

The expression of energy is


For n = 1 ΔE = 3K/4 = 0.75 K
n = 2 ΔE = 5K/36 = 0.14 K 
n = 3 ΔE = 7K/144 = 0.049 K

L_z = m(h/2π) and L_z = L cos θ.  Hence,

The terms within the brackets represents angular dependence.

The function R depends on the quantum numbers n and l, the function Θ depends on l and | m | and the function Ф depends on m.

The transition n₂ =  4 → n₁ = 3 will emit the longest wavelength. Hence

= 1.875 x 10^–6 m = 1875 nm