The 1st- 2nd- and the 3rd ionization enthalpies- I1- I2- and I3 of four atoms with atomic numbers n- n-1- n-2- and n-3- where n-10- are tabulated below-what is the value of n- Atomic Ionisation Enthalpy- Number kJ-mol- I1 I2 I3- n 1681 3374 6050- n-1 2081 3952 6122- n-2 469 4562 6910- n-3 738 1451 7733- -

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Trends in Ionization Energy

The 1st, 2nd,...

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The $1^{s t}, 2^{n d}, and the 3^{r d}$ ionization enthalpies, $I_{1,} I_{2}, and I_{3}$ of four atoms with atomic numbers $n, n + 1, n + 2, and n + 3$ , where $n < 10$ , are tabulated below.
what is the value of $n$ ?
$\begin{matrix} Atomic & Ionisation Enthalpy Number & (k J / m o l) I_{1} & I_{2} & I_{3} n & 1681 & 3374 & 6050 n + 1 & 2081 & 3952 & 6122 n + 2 & 469 & 4562 & 6910 n + 3 & 738 & 1451 & 7733 \end{matrix}$

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I_{1} = n \int 0 [x] d x

I_{2} = n \int 0 {x} d x,

[x]

{x}

x

n \in N - {1}

\frac{I_{1}}{I_{2}}

z = \frac{(1 + i) (1 + 2 i) (1 + 3 i) \dots \dots \dots (1 + n i)}{(1 - i) (2 - i) (3 - i) \dots \dots \dots (n - i)}

i = \sqrt{- 1}, n \in N

z

I_{n} = \int_{0}^{1} \frac{(l n x)^{n}}{\sqrt{x}} d x, n ϵ N,

z = \frac{(1 + i) (1 + 2 i) (1 + 3 i) \dots \dots \dots (1 + n i)}{(1 - i) (2 - i) (3 - i) \dots \dots \dots (n - i)}

i = \sqrt{- 1}, n \in N

z

I_{1}, I_{2}, I_{3}

m

I_{1}

n

I_{2}

k

I_{3}

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