Question

A hydrogen like system has ionization energy 11808 kJ/mol. Find the number of protons in nucleus of system​

Answer 1

Answer: The number of protons that the element contains is 3.

Explanation:

Ionization energy is defined as the energy which is released when a valence electron is ejected out of an isolated gaseous atom.

The formula used to calculate ionization energy follows:

$E_n=\frac{-1312\times Z^2}{n^2}kJ/mol$

where,

Z = atomic number

n = Energy level

The transition of ionization energy is from $1\rightarrow \infty$

We are given:

$E_n=11808kJ/mol$

Putting values in above equation, we get:

$11808=(\frac{-1312\times Z^2}{\infty ^2})-(\frac{-1312\times Z^2}{1^2})\\\\11808=1312\times Z^2\\\\Z=3$

Atomic number is defined as the number of protons that are present in an atom. Atomic number of the element is coming out to be 3 which means that the number of protons that are contained in this atom are 3.

Answer 2

The number of proton are 3

Explanation:

Ionization energy is defined as the energy which is released when a valence electron is ejected out of an isolated gaseous atom.

The formula used to calculate ionization energy follows:

En = - 1312 x z^2 / n^2 K J / mol

Where

Z = atomic number

n = Energy level

The transition of ionization energy is from

We are given:

En = 11808 KJ/ mol

Putting values in above equation, we get:

11808 = (-1313 x z^2 / ∞^2 ) - ( - 1312 x z^2 / 1^2)

11808 = 1312 x z^2

Z = 3

Hence the number of proton are 3