Question

What is the ionization potential of this element?

• A. $3.96V$
• B. $9.23V$
• C. $6.34V$
• D. $8.28V$

Answer 1

The correct option is D $8.28V$

Ionization potential is energy required to excite electron from ground state to $p=\infty$
$E=\frac{hc}{\lambda }$

For, $p=\infty ,\lambda =1500\mathring{A}$
Therefore, Energy required is, $E=\frac{(6.625\times {10}^{-34})(3\times {10}^8)}{1.5\times {10}^{-7}}=13.25\times {10}^{-19}J$$=\frac{13.25\times {10}^{-19}}{1.602\times {10}^{-19}}=8.27eV$