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Saturday, December 8, 2012

9th Class NCERT Solution Chapter: Structure of Atom

Questions:  Explain with examples (i) atomic number (ii) mass number (iii) isotopes and (iv) isobars.  Give any two uses of isotopes.

Ans. (i) Atomic number : The number of protons in the nucleus of an atom is called its atomic number.
Example : Chlorine has 17 protons in its nucleus, therefore, its atomic number is 17.

(ii) Mass number : The number of protons and neutrons present in the nucleus of an atom is known as its mass number.
Example : Sodium has 11 protons and 12 neutrons in its nucleus, therefore, its mass number is (11 + 12) = 23.

(iii) Isotopes : Atoms of the same element having the same atomic number but different mass numbers are called isotopes.
Example : Chlorine has two isotopes 3517Cl and 3717 Cl , in which atomic number is 17 for both, but mass numbers are 35 and 37.

(iv) Isobars : Atoms of different elements having the same mass number, but different atomic numbers are known as isobars.
Example : Calcium and argon atoms have the same mass number 40, but different atomic numbers 20 and 18 respectively.
Uses of Isotopes
1. Isotope of uranium 23592U is used as a nuclear fuel in atomic reactors.
2. Cobalt – 60, an isotope of cobalt is used in the treatment of cancer.
3.Isotopes of iodine is used to treat goitor

Questions: Na+ has completely filled K and L shells. Explain.
Ans. Na loose 1 electron to form sodium ion Na+ . So, Na+ ion has one electron less than Na atom i.e., it has 10 electrons.  Therefore, Electronic configuration of Na+ = 2,8  completely filled k(2) and(8) l shells

Questions: If bromine atom is available in the form of, say, two isotopes 7935 Br (49.7%) and 8135 Br (50.3%), calculate the average atomic mass of bromine atom.
Ans:  The average atomic mass of bromine atom = 49.7% of 79 + 50.3% of 81= 39.263+40.743 = 80.006 u = 80 u,

Questions: The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes 16 8 X and 188 X in the sample?
Ans. Let the percentage of isotope 16 8 X be  y%. then,  the percentage of isotope 188 X  will be (100 − y) %.
Therefore,  16.2 =  16 x y/100 + 18 x (100-y)/100
16.2 =  16y/100 + (1800 - 100y)/100
16.2 x100 = 16y+1800 - 18y
1620 -1800 = -2y
-180/-2 = y = 90
So, 16 8 X = 90%  and  18 8 X     = 10%
Questions: If Z = 3, what would be the valency of the element? Also, name the element.
Ans: Z = 3, we mean that the atomic number of the element is 3. Its electronic configuration is 2, 1. Hence, the valency of the element is 1 Therefore, the element with Z = 3 is lithium.

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