Question

If you are given Avogadro's number of atoms of a gas 'X'. If half of the atoms are converted into $X_{\left(g\right)}^{+}$ by energy $\Delta H$. The IE of X is

• A. $\frac{2\Delta H}{N_A}$
• B. $\frac{2N_A}{\Delta H}$
• C. $\frac{\Delta H}{2N_A}$
• D. $\frac{N_A}{\Delta H}$

Answer 1

The correct option is A $\frac{2\Delta H}{N_A}$

Given number of atoms=Avogadro's number of atoms $=N_A=6.023\times {10}^{23}$
Given that $\frac{N_A}{2}$ atoms are ionised, i.e.,
Ionsation energy of $\frac{N_A}{2}$ atoms of gas $X=\Delta H$
$\therefore$ Ionisation energy of one atom of gas $X=\frac{\Delta H}{\left(\frac{N_A}{2}\right)}=\frac{2\times \Delta H}{N_A}$
Hence, ionisation energy of gas X is $\frac{2\times \Delta H}{N_A}$