JEE Main2014ChemistryStructure of AtomActual
The 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 × 1 0 - 3 4 Js , N A = 6.022 × 1 0 2 3 mol -1 )
Options
- A1.24 × 1 0 1 5   s -1
- B7.50 × 1 0 4   s -1
- C4.76 × 1 0 1 4   s -1
- D3.15 × 1 0 1 5   s -1
Correct answer
A. 1.24 × 1 0 1 5   s -1
Step-by-step solution
One photon ionises one Na - atom by elastic collision. NA     hv photon × 10 – 3   ( KJ   mole – 1 ) 6 · 022 × 10 23 × 6 · 626 × 10 - 34 × υ photon × 10 - 3 = 495 · 5 υ photon = 495 · 5 6 · 022 × 6 · 626 × 10 - 14 = 12 . 41 × 10 14 = 1 · 2 4 1 × 1 0 1 5   sec - 1