Question

Ionisation energy of lithium atom is $5.2 \times 10^{2} \,kJ\, mol ^{-1}$. The wavelength of radiation sufficient to ionise sodium atom is

• A. $0.0289\, nm$
• B. $0.229\, nm$
• C. $229\, nm$
• D. $289\, nm$

Answer 1

Answer: (C)

$I . E .=5.2 \times 10^{2} kJ mol ^{-1}=\frac{5.2 \times 10^{5}}{6.023 \times 10^{23}} J\, atom ^{-1}$

But $E=\frac{h c}{\lambda}$

$\therefore \,\,\,\, \lambda=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8} \times 6.023 \times 10^{23}}{5.2 \times 10^{5}}$

$=229 \times 10^{-9} m =229 \,nm$