Complete the following table.
ELEMENT
$n$
$p$
$e$
MASS NUMBER
ATOMIC NUMBER
$Be$
_
4
_
9
_
$Mg$
12
_
12
_
_
$Si$
14
_
_
_
14
$P$
_
15
_
31
_

ELEMENT	$n$	$p$	$e$	MASS NUMBER	ATOMIC NUMBER
$Be$	_	4	_	9	_
$Mg$	12	_	12	_	_
$Si$	14	_	_	_	14
$P$	_	15	_	31	_

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Solution

BERYLLIUM ( $B e$ )
Given, No. of protons, $p = 4$ and Mass Number $= 9$
$A t o m i c n u m b e r = N o . o f p r o t o n s$ , Therefore atomic number $= 4$
$N o . o f p r o t o n s = N o . o f e l e c t r o n s$ , Therefore $e = 4$
$M a s s N u m b e r = p r o t o n s + n e u t r o n s$ , Therefore $N o . o f n e u t r o n s =$ $M a s s N u m b e r - p r o t o n s$ $= 9 - 4$
Hence, $n$ $= 5$
MAGNESIUM ( $M g$ )
Given, No. of neutrons, $n = 12$ and No. of electrons, $e = 12$
$A t o m i c n u m b e r = N o . o f p r o t o n s$ , Therefore atomic number $= 12$
$N o . o f p r o t o n s = N o . o f e l e c t r o n s$ , Therefore, $p = 12$
$M a s s N u m b e r = p r o t o n s + n e u t r o n s$ $= 12 + 12$ $= 24$
SILICON ( $S i$ )
Given, No. of neutrons, $n = 14$ and atomic number $= 14$
$A t o m i c n u m b e r = N o . o f p r o t o n s$ , therefore, No. of protons, $p = 14$
$N o . o f p r o t o n s = N o . o f e l e c t r o n s$ , therefore, No. of electrons, $e = 14$
$M a s s N u m b e r = p r o t o n s + n e u t r o n s$ $= 14 + 14 = 28$
PHOSPHORUS ( $P$ )
Given, No. of protons, $p = 15$ and Mass Number $= 31$
$N o . o f p r o t o n s = N o . o f e l e c t r o n s$ , therefore no. of electrons, $e = 15$
$A t o m i c n u m b e r = N o . o f p r o t o n s$ , therefore atomic number $= 15$
$M a s s N u m b e r = p r o t o n s + n e u t r o n s$ , therefore $no . of neutrons = mass number - no . of protons$ $= 31 - 15 = 16$
ELEMENT
$n$
$p$
$e$
MASS NUMBER
ATOMIC NUMBER
$Be$
5
4
4
9
4
$Mg$
12
12
12
24
12
$Si$
14
14
14
28
14
$P$
16
15
15
31
15

ELEMENT	$n$	$p$	$e$	MASS NUMBER	ATOMIC NUMBER
$Be$	5	4	4	9	4
$Mg$	12	12	12	24	12
$Si$	14	14	14	28	14
$P$	16	15	15	31	15

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