Solved Examples on Structure of Atom

Atoms are the basic building blocks of matter which has mass and occupies space. They cannot be chemically subdivided byordinary means. The word ‘atom’ is derived from the Greek word atom which means indivisible. Matter can be broken into small particles called atoms which are too small to be seen through the naked eye.

As discussed earlier, atoms are composed of three types of particles namely protons, neutrons and electrons. A major proportion of mass of an atom is due to protons and neutrons while electrons have very small mass of about (9.108 X 10-28 grams).

We have enlisted the information of various particles along with their charges in the table given below

Particle

Charge

Mass (g)

Mass (amu)

Proton

+1

1.6727 x 10-24 g

1.007316

Neutron

0

1.6750 x 10-24 g

1.008701

Electron

-1

9.110 x 10-28 g

0.000549

 

We discuss some of the illustrations on the structure of an atom:

Illustration:A doubly ionized Lithium atom is hydrogen like with atomic number 3.

(i) Find the wavelength of radiation required to excite the electron in Li++ from the first to the third Bohr Orbit. (Ionization energy of the hydrogen atom equals 13.6 eV).

(ii) How many spectral lines are observed in the emission spectrum of the above excited system?

Solution:We are required to find theenergy required to excite doubly ionized lithium. We know 

En = –13.6 Z2 / n2

Hence, the excitation energy = ΔE = E3 – E1 = –13.6 × (3)2 [1/32 – 1/12]

= +13.6 × (9) [1–1/9] = 13.6 × (9) (8/9) = 108.8 eV.

Wavelength λ = hc / ΔE = (6.63 × 10–34) (3×108) / (13.6 × 8) (1.6 × 10–19)

                = (6.63 × (3) / (13.6) (8)(1.16)) 10–7 

                = 114.26 × 10–10 m 

                = 1143 A

(ii) From the excited state (E3), coming back to ground state, there can be 3C2 = 3 possible radiations.

View this video for more examples on structure of atom

Illustration:

Radiation emitted in the transition n = 3 to n = 2 orbit in a hydrogen atom falls on a metal to produce photoelectrons, the electrons emitted from the metal surface with the maximum kinetic energy are allowed to pass through a magnetic field of strength 3.125x10–4 T and it is observed that the paths have radius of curvature as 10 mm. Find

(a)  The kinetic energy of the fastest photo electrons.

(b) The work function of the metal.

(c)  The wavelength of the radiation.

Solution:First we discuss how to find the kinetic energy of the fastest photoelectron. The formula used for finding the same is

ΔE = hv23 – 13.6 (1/22 – 1/32) eV

The Kinetic energy of the fastest photoelectrons is given by

       mv2 / r = Bev 

or     mv = p = Ber

Kinetic Energy = 1/2 mv2 = B2e2r2 / 2m

                = (3.125 × 10–4)2 × (1.6 ×)

                = 0.86 eV

(a) The kinetic energy of the fastest electron is 0.86 eV

(b) The work function, Φ is given by

            Φ = (1.9 – 0.86) eV = 1.04 eV

(c)  The wavelength of the emitted radiation is,

            λ = hc/E = (6.63 × 10–34) × (3×10–8) / 1.3 × 1.6 × 1–19 m

               = 6.54 × 10–7 m

                       = 6540Å

Atoms are the basic unit of matter and it is vital to have a thorough knowledge of its components like electrons, protons and neutrons. Various questions are often asked in JEE on the computation of wavelength of the fastest photoelectrons emitted or the calculation of atomic number. It is important for the JEE aspirants to master these topics in order to remain competi9tive in the JEE.

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