Electromagnetic radiation of wavelength 663 nm is just sufficient to ionise the atom of metal A. The ionization energy of metal A in kJ mol⁻¹ is ________ . (Rounded-off to the nearest integer) [h=6.63$\times$10⁻³⁴ Js, c=3.00$\times$10⁸ ms⁻¹, N$_A$=6.02$\times$10²³ mol⁻¹]
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A useful shortcut for energy calculations is $E(\text{kJ/mol}) = \frac{1.196 \times 10^5}{\lambda(\text{nm})}$. Using this, $E = \frac{119600}{663} \approx 180.4$ kJ/mol, which rounds to 180 or 181 depending on precision. The direct calculation is more reliable.
Step 1: Concept used
When radiation is just sufficient to ionise an atom, the energy of one photon is equal to the ionisation energy of one atom.
The energy of a photon is given by:
\[
E = \frac{hc}{\lambda}
\]
Step 2: Substitute given values
\[
h = 6.63 \times 10^{-34}\ \text{J s}
\]
\[
c = 3.00 \times 10^{8}\ \text{m s}^{-1}
\]
\[
\lambda = 663\ \text{nm} = 663 \times 10^{-9}\ \text{m}
\]
Step 3: Calculate energy per atom
\[
E_{\text{atom}} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}
{663 \times 10^{-9}}
\]
\[
E_{\text{atom}} = \frac{19.89 \times 10^{-26}}{6.63 \times 10^{-7}}
\]
\[
E_{\text{atom}} \approx 3.00 \times 10^{-19}\ \text{J}
\]
Step 4: Convert energy per atom to energy per mole
\[
N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}
\]
\[
E_{\text{mole}} = E_{\text{atom}} \times N_A
\]
\[
E_{\text{mole}} = (3.00 \times 10^{-19}) \times (6.02 \times 10^{23})
\]
\[
E_{\text{mole}} = 18.06 \times 10^{4}\ \text{J mol}^{-1}
\]
Step 5: Convert J/mol to kJ/mol
\[
E_{\text{mole}} = \frac{180600}{1000} = 180.6\ \text{kJ mol}^{-1}
\]
Step 6: Final Answer
Rounding off to the nearest integer:
\[
\boxed{181\ \text{kJ mol}^{-1}}
\]
