Ionization energy of gaseous

Question:

Ionization energy of gaseous $\mathrm{Na}$ atoms is $495.5 \mathrm{kjmol}^{-1}$. The lowest possible frequency of light that ionizes a sodium atom is $\left(\mathrm{h}=6.626 \times 10^{-34} \mathrm{Js}, \mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1}\right)$

  1. $3.15 \times 10^{15} \mathrm{~s}^{-1}$

  2. $4.76 \times 10^{14} \mathrm{~s}^{-1}$

  3. $1.24 \times 10^{15} \mathrm{~s}^{-1}$

  4. $7.50 \times 10^{4} \mathrm{~s}^{-1}$

Correct Option: , 3

Solution:

Leave a comment

eSaral Gurukul

NEET

JEE Main

JEE Advanced

Our Telegram Channel

All Study Material