Answer:
The frequency is [tex]f = 2,01 * 10^{15} \ Hz [/tex]
Explanation:
From the question we are told that
The energy required to ionize boron is [tex]E_b = 801 KJ/mol[/tex]
Generally the ionization energy of boron pre atom is mathematically represented as
[tex]E_a = \frac{E_b}{N_A}[/tex]
Here [tex]N_A[/tex] is the Avogadro's constant with value [tex]N_A = 6.022*10^{23}[/tex]
So
[tex]E_a = \frac{801}{6.022*10^{23}}[/tex]
=> [tex]E_a = 1.330 *10^{-18} \ J/atom [/tex]
Generally the energy required to liberate one electron from an atom is equivalent to the ionization energy per atom and this mathematically represented as
[tex]E = hf = E_a[/tex]
=> [tex] hf = E_a[/tex]
Here h is the Planks constant with value [tex]h = 6.626 *10^{-34}[/tex]
So
[tex]f = \frac{1.330 *10^{-18}}{ 6.626 *10^{-34}}[/tex]
=> [tex]f = 2,01 * 10^{15} \ Hz [/tex]
