JEE Mains · Chemistry · STD 11 - 2. structure of atom
Ionization energy of gaseous \(Na\) atoms is \(495.5\, kJ\, mol^{-1}\) . The lowest possible frequency of light that ionizes a sodium atom is \((h\, = 6.626\times10^{-34}\, Js, N_A\, \)\(= 6.022\times10^{23}\, mol^{-1} )\)
- A \(7.50\times10^4\, s^{-1}\)
- B \(4.76\times10^{14}\, s^{-1}\)
- C \(3.15\times10^{15}\, s^{-1}\)
- D \(1.24\times10^{15}\, s^{-1}\)
Answer & Solution
Correct Answer
(D) \(1.24\times10^{15}\, s^{-1}\)
Step-by-step Solution
Detailed explanation
Energy \(= N_A\,hv\) \(495.5 = 6.023 \times 10^{23} \times 6.6 \times 10^{-34} \times v\) \(v = \frac{{495.5 \times {{10}^3}\,J}}{{6.023 \times {{10}^{23}} \times 6.6 \times {{10}^{ - 34}}}} = 12.4 \times {10^{14}}\) \( = 1.24 \times {10^{15}}\,{s^{ - 1}}\)
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