Question

The energy required to ionize boron is 801 kJ/mol. You may want to reference (Pages 93 - 98) Section 2.5 while completing this problem. Part A What minimum frequency of light is required to ionize boron

Answer 1

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

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

=>      $f = 2,01 * 10^{15} \ Hz$