# First ionization energy of two elements X and Y are 500 kg/mol and 375 kg/mol . (kg/mol is unit of ionization energy).Comment about their relative position if they are located in the same i) Same group ii)Same period

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by ltv257

## Answers

2015-05-25T11:51:13+05:30

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Z = atomic number            n = quantum number

Ionization energy E = Z² e² / [ 2 n² r₀ ]
r₀ = h² / (4 π² m e²)  = First Bohr radius
E =  13.6 Z² / n²  eV    =  21.787  Z² / n²  Joules

Ionization energy increases  in a period and decreases in a group.

n decreases down in a group.    Z increases in a period to the right.

1) If they are in the same group:
Y is below X.

Z1 = Z2 + 2  or  Z2 + 8  or  Z2 + 18 , Z2 + 10, or Z2 + 16 or Z2 + 26 or Z2 + 36
or  Z2 + 34  or Z2 + 32 or Z2 + 68 or Z2 + 84  or Z2 + 86  or Z2 + 50
n1 = n2 + 1, n2 + 3 or n2 + 4 or n2 + 5

If  500 kJ / 375  =  (Z2 + x) / Z2,         Let Z1 = Z2 + x
=>  1 +  x /Z2  =  1.154
=>   x/Z2 = 0.154  => Z2  = 6.49 x  ≈ 6.5 x

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2) If they are in the  same period:
then  X is to the right of  Y.

Z1  =  Z2 + 1 or Z2  +  2 ,... Z2 + 7,..., or  Z2 + 17
n is the same.

E1 / E2  =  500 / 375  =  Z1² / Z2²
=>    Z1 / Z2   ≈ 1.154  ≈  23 / 20      or  46 / 40
=>    So  Z1 = 23    and    Z2 = 20
or  Z1 = 46  and  Z2 = 40
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