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makvit [3.9K]
2 years ago
13

The energy required to ionize boron is 801 kJ/mol. You may want to reference (Pages 93 - 98) Section 2.5 while completing this p

roblem. Part A What minimum frequency of light is required to ionize boron
Chemistry
1 answer:
earnstyle [38]2 years ago
8 0

Answer:

The frequency is  f =  2,01 * 10^{15} \  Hz

Explanation:

From the question we are told that

   The energy required to ionize boron is E_b  =  801 KJ/mol

Generally the ionization energy of boron pre atom is mathematically represented as

     E_a  =  \frac{E_b}{N_A}

Here  N_A is the Avogadro's constant with value N_A  =  6.022*10^{23}

So

      E_a  =  \frac{801}{6.022*10^{23}}

=>     E_a  =  1.330 *10^{-18} \  J/atom

Generally the energy required to liberate one electron from an atom is equivalent to the ionization energy per atom and this mathematically represented as

       E =  hf  =  E_a

=>     hf  =  E_a

Here h is the Planks constant with value h = 6.626 *10^{-34}

So

       f =  \frac{1.330 *10^{-18}}{ 6.626 *10^{-34}}

=>      f =  2,01 * 10^{15} \  Hz

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