Home 1st Year Chemistry Notes FSC 1st Year Chemistry Chapter 5 Solved Exercise Notes PDF

FSC 1st Year Chemistry Chapter 5 Solved Exercise Notes PDF


In this post, I am sharing FSC 1st Year Chemistry Chapter 5 Solved Exercise PDF Notes for the students of FSC Part 1. The name of 11th Chemistry Chapter 11 is Atomic structure. So the students can download Atomic structure Chapter Solved Exercise in PDF format from here. This file contains 29 pages. These Chemistry Notes are for all the boards working under Punjab Board like Gujranwala Board, Lahore Board, Faisalabad Board, Multan Board, Rawalpindi Board, Sargodha Board, DG Kahn Board, and Sahiwal Board. Here are the complete 1st Year Chemistry Notes.

11th Class Chemistry Chapter 5 “Atomic Structure” Solved Exercise Notes PDF Download

The nature of positive rays depends on;

a) the nature of electrode
b) the nature of discharge tube
c) the nature of residual gas
d) all of the above

The velocity of the photon is;

a) independent of its wavelength
b) depend on its wavelength
c) equal to square of its amplitude
d) depends on its source

Rutherford’s model of atom failed because

a) the atom did not have a nucleus and electrons
b) it did not account for the attraction between proton and neutron
c) it did not account for the stability of the atom
d) there is actually no space between the nucleus and electrons

In the ground stat of an atom, the electron is present:

a) in the nucleus
b) in the second shell
c) nearest to the nucleus
d) farthest from the nucleus

Fill in the blanks:

The charge on one mole of electron is 96484C coulambs.
The mass of hydrogen atom is 1.66 x 10-24
energy is released when electron jumps from higher to a lower orbit.
the number of electrons in a given subshell is given by formula 2

Justify the distance gaps between different orbits go on increasing from lower to the higher orbit.

For hydrogen atom we can calculate radius of different orbits and their differences as follows

                                    rn = 0.529(n2/Z) Ao

For H,Z =1 and

                                                rn = 0.529Ao x n2
now if

                        n = 1                r1 = 0.529Ao
                        n =2                 r2 = 0.529x 4 = 2.11 Ao
                        n =3                 r3 =0.529 x 9 = 4.76 Ao
                        n= 4`                r= 0.529 x 16 =8.46 Ao

and the difference are;
                        r– r1   =          2.11 – 0.529 = 1.58 Ao

                        r– r2    =          4.76 – 2.11 = 2.65 Ao

                        r– r3    =          8.46 – 4.76 = 3.7 Ao

these value shown that gaps between adjacent orbits increases when we move from lower to higher orbits.

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