Atoms class 12 NCERT solutions explain the Rutherford model, wherein he performed an experiment in which he bombarded a thin sheet of gold with -particles and then tracked the trajectory of these particles as they collided with the gold foil.

According to the Rutherford atomic model:

• Positively charged particles and the majority of an atom's mass are concentrated in a small volume called the nucleus.
• The Rutherford model also proposed that the nucleus is surrounded on all sides by negatively charged electrons. Electrons that surround the nucleus travel in circular paths (orbits) at great speeds.
• The nucleus is held together by a strong electrostatic force of attraction. 

Limitations

• Rutherford’s model could not explain the stability of an atom.

When alpha particles are bombarded with a certain initial velocity, the multiple trajectories that the alpha particle takes are observed. The impact parameter determines the trajectory of an alpha particle and is defined as the perpendicular distance between the initial velocity vector and the centre of the target nucleus. The impact parameter ‘b’ is inversely proportional to the initial kinetic energy of the alpha particles.

According to NCERT solutions of atoms class 12, the centripetal force (Fc) required to keep the revolving electrons in their orbits is provided by the electrostatic force of attraction, Fe, between the revolving electrons and the nucleus. Thus, for a dynamically stable orbit in a hydrogen atom,

Fe = Fc 

When an electron transitions from one energy level to another, it either emits or absorbs light of a specific wavelength. The spectrum of the electromagnetic radiation that is emitted or absorbed by electrons while transitioning between different energy levels within an atom is called atomic spectra.

RYDBERG’S FORMULA:

1/λ=RZ2(1/n′2-1/n2 )

λ is the wavelength, R is Rydberg constant, and n may have integral values 3, 4, 5, etc. R is 1.097 × 107 m–1.

Z   atomic number,

n  upper energy level,

n' lower energy level

A photon of energy n′ – n is released when electrons jump from higher energy states (n′) to lower energy ones (n). As the energy corresponding to each state is fixed, the difference between them is also fixed. As a result, the transition between similar energy states will produce photons of the same energy.

For hydrogen atom,

Lyman series: 1/λ=R(1/12 -1/n2) n= 2,3,4... 

Balmer series: 1/λ=R(1/22 -1/n2) n = 3,4,5,6... 

paschen series: 1/λ=R(1/32 -1/n2) n = 4,5,6,7... 

Brackett series: 1/λ=R(1/42 -1/n2) n = 5,6,7,8...

Pfund series: 1/λ=R(1/52 -1/n2) n = 6,7,8,9...

• An electron in an atom revolves in certain stable orbits without the emission of radiant energy.
• The electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2π, where h is the Planck’s constant (= 6.6 × 10–34 J s). Thus, the angular momentum (L) of the orbiting electron is quantised.

L = nh/2π

• The frequency of the emitted photon hν = Ei – Ef (12.12)

Ei and Ef are the energies of the initial and final states and Ei> Ef.

Bohr’s Equation

r(n)=n2×r(1)

• n is a positive integer
• r(1), also known as Bohr’s radius, is the smallest allowed radius for the hydrogen atom

r(1)=0.529×10−10m

The energy of an electron in the nth level of hydrogen by considering the electrons in circular and quantised orbits as:

E(n)=−1/n2×13.6 eV

According to Bohr’s model, an electron absorbs energy in the form of photons to get excited to a higher energy level. Because the excited electron is less stable after escaping to a higher energy level, also known as the excited state, it quickly emits a photon to return to a lower, more stable energy level. The difference in energy between the two energy levels for a particular transition is equal to the energy of the emitted photon.

hv=ΔE=(1/n’2−1/n2)13.6eV

According to atoms class 12 NCERT solutions, when an atom's electron is revolving in the orbit closest to the nucleus, n = 1, the energy of the atom is the least (biggest negative value). The absolute value of the energy E is smaller for n = 2, 3,..., hence, the energy is progressively larger in the outer orbits. The ground state of an atom is the state with the lowest energy. Here, the electron circles in the smallest radius orbit, the Bohr radius, a0. This state (n = 1) has an energy of –13.6 eV. As a result, the minimum energy necessary to release one electron from the hydrogen atom's ground state is 13.6 eV. The ionisation energy of the hydrogen atom is what it is termed as.

According to atoms class 12 NCERT solutions, the third postulate of Bohr’s model, when an atom makes a transition from a higher energy state with quantum number of ni to a lower energy state with quantum number nf (nf < ni ), the difference of energy is carried away by a photon of frequency νif such that

hνif = Eni – Enf

When electrons in an atom or molecule absorb energy and become excited, they jump from a lower to a higher energy level, and when they return to their original states, they emit radiation. This phenomenon occurring in hydrogen atoms is called hydrogen emission spectrum.

• Transition from the first shell to any other shell – Lyman series
• Transition from the second shell to any other shell – Balmer series
• Transition from the third shell to any other shell – Paschen series
• Transition from the fourth shell to any other shell – Bracket series
• Transition from the fifth shell to any other shell – Pfund series

v= 109,677 (1/n′2-1/n2)

Where,

n′ = 1,2,3,4 …

n =n′ +1

ν= wavenumber of electromagnetic radiation. The value 109,677 cm-1 is known as the Rydberg constant for hydrogen.

The circumference of the circular orbit must be an integral number of wavelengths:

2πr=nλ=nh/p (λ=h/p)

The momentum, p, is simply mv as long as we're talking about non-relativistic speeds.

Therefore, this becomes:

2πr=nh/mv

Rearranging Lr=mvr=nh/2π

A 10 kg satellite circles earth once every 2 h in an orbit having a radius of 8000 km. Let’s assume that Bohr’s angular momentum postulate applies to satellites just as it does to an electron in the hydrogen atom. Find the quantum number of the orbit of the satellite.

Solution:

m = 10 kg and rn = 8 × 106 m                    T = 7200 s.

vn = 2π rn/T.

mvr=nh/2π

n = (2π rn) 2 × m/(T × h).

n = (2π × 8 × 106 m)2 × 10/(7200 s × 6.64 × 10–34 J s)= 5.3 × 1045

1. What are atomic spectra according to Chapter 12 of NCERT Solutions for Class 12 Physics?

The atomic spectra is the spectrum of the electromagnetic radiation that is emitted or absorbed by electrons while transitioning between different energy levels within an atom.

2. What sort of questions does NCERT Solutions Class 12 Physics offer for chapter 12?

Physics atoms class 12 NCERT solutions offer competitive-based questions which are associated with concepts from other chapters like electricity and electromagnetic induction.

3. What is the structure of atoms?

The protons (positively charged) and neutrons are found in the nucleus (centre) of the atom (no charge). The electrons are contained in the electron shells (negatively charged).

4. Is everything made out of atoms?

Yes, everything in the universe is made of atoms. Each atom is made up of subatomic particles.