Question

Ionization energy for hydrogen atom in ergs, Joules and eV respectively is :

• A. $21.8\times {10}^{-12},218\times {10}^{-20},13.6$
• B. $13.6\times 218\times {10}^{-20},21.8\times {10}^{-13}$
• C. $21.8\times {10}^{-20},13.6,21.8\times {10}^{-13}$
• D. $21.8\times {10}^{-13},13.6,21.8\times {10}^{-20}$

Answer 1

Answer: (A)

$21.8\times {10}^{-12},218\times {10}^{-20},13.6$

$\Delta E=E_{\infty }-E_1$

$\Delta E=0-(-13.6\times z²/n²)eV$

In this case we have hydrogen atom therefore Z =1 and N =1
$\Delta E=13.6eV$
now $1eV=1.6\times {10}^{-19}$ joules
therefore $\Delta E=13.6\times 1.6\times {10}^{19}$
$\Delta E=218\times {10}^{-20}$ joules

now $1joule={10}^7$ ergs

$\Delta E=218\times {10}^{-}20\times {10}^7$ ergs

$\Delta E=21.8\times {10}^{-12}$ ergs

Hence option A is correct