Question

A hydrogen like ion is an ion containing only one electron. The energies of the electron in a hydrogen like ion are given by $E_n=-(2.18\times {10}^{-18})Z^2\left(\frac{1}{n^2}\right)$ joules, where, n is the principal quantum number and Z is the atomic number of the element. Then the ionisation energy (in kJ mol${}^{-1}$) of the He${}^{+}$ ion is:

• A. $2.752\times {10}^{3}kJmo{l}^{-1}$
• B. $5.2494\times {10}^{3}kJmo{l}^{-1}$
• C. $4.5214\times {10}^{3}kJmo{l}^{-1}$
• D. $7.482\times {10}^{3}kJmo{l}^{-1}$

Answer 1

The correct option is B $5.2494\times {10}^{3}kJmo{l}^{-1}$

$(IE{)}_{H{e}^{+}}=(-E_n)$$=\frac{2.18\times {10}^{-18}{Z}^2}{n^2}Jato{m}^{-1}$

$=\frac{2.18\times {10}^{-18}\times (2{)}^2}{1}Jato{m}^{-1}$ for $H{e}^{+}(n=1,Z=2)$

$=2.18\times {10}^{-18}\times 4\times 6.02\times {10}^{23}Jmo{l}^{-1}$

$=5.2494\times {10}^{5}Jmo{l}^{-1}$

$=5.2494\times {10}^{3}kJmo{l}^{-1}$

Hence, the correct option is $\text{B}$