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Question

(a)Atomic number of an atom is 16 and the mass number is 32
(i) Find the number of protons, electrons and neutrons in the atom.
(ii) Write the electronic configuration of the atom.
[2 Marks]
(b) Which of the following element is isotopic in nature.

19Y40 , 20Z40 , 19X41, 19U39 [1 Mark]

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Solution

(a)
(i) Atomic number = 16
Mass number = 32
We know that atomic number is equal to number of proton and number of proton is equal to the number of electron.
So, Proton number = electron number = 16.
We also know that,
Mass number = number of neutrons + number of proton
32 = X + 16 ( X= number of neutron)
X=16 [1 Mark]

(ii) The electronic configuration of the element is 2,8,6.
[1 Mark]

(b) Isotopes are members of a family of an element that all have the same number of protons but different numbers of neutrons. Out of these four elements, element Y, X, and U have same atomic number or proton number and different mass number. So, 19Y40 , 19X41, and 19U39 are isotopic in nature. [1 Mark]

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