An element may be represented as ${}_{18}^{40}X$ . Find the number of electrons, protons and neutrons in this atom. Give its electronic configuration.

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Solution

The given element is of the form ${}_{atomic number}^{mass number}X$ ( ${}_{18}^{40}X$
Therefore mass number of the element = $40$ and atomic number = $18$
$Atomic number = no . of protons$ , therefore no. of protons = $18$
$no . of protons = no . of electrons$ , therefore no. of electrons = $18$
Since $mass number = no . of protons + no . of neutrons$ ,
$no . of neutrons = mass number - no . of protons$
Therefore no. of neutrons = $40 - 18 = 22$
Hence,
e= $18$
p= $18$
n= $22$
Electronic configuration = $K L M 2 8 8$ OR $2, 8, 8$

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An element may be represented as X1840. Find the number of electrons, protons and neutrons in this atom. Give its electronic configuration.

An element may be represented as ${}_{18}^{40}X$ . Find the number of electrons, protons and neutrons in this atom. Give its electronic configuration.