Question

Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its electrons and therefore would exist as $H{g}^{80+}$. Energy required for the step is:

$H{g}^{79+}\left(g\right)\to H{g}^{80+}\left(g\right)+e^{-}$

• A. $4.095\times {10}^{6}kJmo{l}^{-1}$
• B. $6.25\times {10}^{6}kJmo{l}^{-1}$
• C. $8.39\times {10}^{6}kJmo{l}^{-1}$
• D. $4.43\times {10}^{6}kJmo{l}^{-1}$

Answer 1

Answer: (C)

$8.39\times {10}^{6}kJmo{l}^{-1}$

$H{g}^{79+}\left(g\right)$ is isoelectronic of hydrogen (electron $=$ 1) $(Z=80,n=1)$

$\therefore (IE{)}_{H{g}^{79+}}=\frac{2.18\times {10}^{-18}\times (80{)}^2}{1^2}Jato{m}^{-1}$

$=1.395\times {10}^{-14}Jato{m}^{-1}$

$=\frac{2.18\times {10}^{-18}\times (80{)}^2\times 6.02\times {10}^{23}}{1000}kJmo{l}^{-1}$

$=8.39\times {10}^{6}kJmo{l}^{-1}$