rydberg constant value in metre , definition , formula class 12 , what is examples

find what is the rydberg constant value in metre , definition , formula class 12 , examples ?

Rutherford’s a-particle Scattering Experiment

A beam of fast a-particles was made to fall on a thin metal foil. Rutherford observed that most of the α-particles passed through the foil without any appreciable deflection. The distance of the closest approach of an α-particle is given by

is the initial kinetic energy of particle and Z is the atomic number of the nucleus. The impact parameter is given by

where θ is the scattering angle.

2. Rutherford’s Model of Atom

The observations of the alpha particle scattering experiments led Rutherford to suggest the following model of atom. (i) An atom has a small central core called the nucleus where almost the entire mass and all the positive charge of the atom is concentrated. The size of the nucleus is about 10^-14 m.

(ii) The remaining part of the atom contains electrons which revolve round the nucleus in various orbits. The size of the atom is about 10^-10m.

(iii) The atom as a whole is electrically neutral; the total negative charge of the electrons surrounding the nucleus is equal to the total positive charge of the nucleus.

3. Bohr’s Theory of Hydrogen Like Atom

(a) Bohr’s quantization condition: The magnitude of angular momentum of the electron in a circular orbit is

where m = mass of electron, r_n = radius of nth circular orbit, v_n = orbital speed of electron in the nth orbit, h = Planck’s constant and n is an integer called the principal quantum number.

(b) Speed of electron in nth orbit is

where Z = atomic number of atom. For hydrogen Z = 1. For a given atom v_n ∝ 1/n . Substituting the known values of e, ε₀ and h we get

(c) Radius of nth orbit is

(d) Total energy of electron in nth orbit

Total energy of electron in nth object is

E_n = K.E. + P.E. = K.E. – 2 K.E. = – K.E

(e) Time period of revolution of electron in nth orbit is

(f) Frequency of revolution of the electron in nth orbit is

(g) Wavelength of emitted radiation: when an electron jumps from a higher energy state n = n2 to a lower state n = n1, a photon of energy hv of radiation is emitted.

R_H is called Rydberg constant.

(h) Main Series of Hydrogen Spectrum (Z = 1)

(1) Lyman series: n1 = 1, n2 = 2, 3, 4, … ∞

Spectral lines in Lyman series lie in the ultraviolet region

(2) Balmer series: n1 = 2, n2 = 3, 4, 5, … ∞

Spectral lines in Balmer series lie in the visible region.

(3) Paschen series: n1 = 3, n2 = 4, 5, 6, … ∞

Spectral lines in Paschen series lie in the infrared region.
(4) Brackett series: n1 = 4, n2 = 5, 6, 7, … ∞

These spectral lines also lie in the far infrared region.

(i) The energy of electron in hydrogen atom in the ground state is – 13.6 eV.

(ii) The ionization potential of hydrogen atom in the ground state is 13.6 V. Ionization potential of a hydrogen like atom in nth state =

Example ; How many wavelengths can be emitted when hydrogen atoms excited to n = 4 state return to the ground state? Find the highest wavelength present in the radiation.

Solution The number of possible wavelengths present in the radiation = number of possible transitions from n = 4 state which is equal to

Example 10 A sample contains hydrogen atoms excited to a state n1. Photons of energy 2.86 eV take the atoms to a higher energy state n2. Then
(a) n1 = 1, n2 = 4 (b) n1 = 2, n2 = 3
(c) n1 = 3, n2 = 5 (d) n1 = 2, n2 = 5

Solution When a photon is absorbed by an atom, it delivers its energy to the atom which shifts the electron to a higher energy state. It follows from Example 9 above that a difference of 2.86 eV can only be absorbed in the transition n = 2 to n = 5 because 3.4 – 0.54 = 2.86 eV. Thus the correct choice is (d).