What minimum frequency of light is required to ionize boron?

Answer 1

E = hv ( h = Planck's constant = 6.6 x 10^ -34J,v = frequency)

E = 801 kJ/mol (or) 801 x 1000 J/mol

The enrgy required for one atom = 801 x 1000 / 6.02 x 10^23

v = E / h

v = 801 x 1000 /6.6 x 10^ -34 x 6.02 x 10^23

= 20.2 x 10^14 Hz

Answer 2

The minimum frequency of light required to ionize boron is about 6.9 x 10^15 Hz, corresponding to a wavelength of approximately 434 nm. This frequency, known as the ionization threshold frequency, is the minimum amount of energy needed to remove an electron from a boron atom.