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Ionization energy of gaseous Na atoms is 495.5  kJmol-1.  The  lowest  possible   
frequency of light that ionizes a sodium   atom is  (h= 6.626\times10-34 Js,   NA= 6.022\times1023 mol-1)  

 

  • Option 1)

     7.50 x 104     s-1   
     

     

  • Option 2)

     4.76 x 1014    s-1  
     

  • Option 3)

    3.15 x x1015    s-1  
     

  • Option 4)

    1.24 x 1015    s-1   

 

Answers (1)

best_answer

As we have learnt,

 

The energy (E) of a quantum of radiation -

E=hv

Where h is plank’s constant and \nu is frequency

-

 

 As we know,

\\*\frac{\Delta E}{N_A} = hv \\*v = \frac{\Delta E}{N_A \times h}

    =\frac{495.5\times 10^3}{6.626\times 10^{-34}\times 6.022\times 10^{23}} = 1.24\times 10^{15}s^{-1}

 


Option 1)

 7.50 x 104     s-1   
 

 

Option 2)

 4.76 x 1014    s-1  
 

Option 3)

3.15 x x1015    s-1  
 

Option 4)

1.24 x 1015    s-1   

Posted by

SudhirSol

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