What minimum frequency of light is required to ionize boron?

The energy required to ionize boron is 801 kJ/mol. What minimum frequency of light is required to ionize boron? Please explain how to do this, I ended up getting 1.208*10^36 and it ended up being wrong. I can’t seem to figure it out, please help!

2 Answers

  • 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

  • Is light strong enough to ionize boron?

    The kind of energy you can get from light

    E=mc^2; c= velocity of light. I don’t know your wavelength, but the only way you can get the frequency with the variables you have put up (energy, frequency) is to use the formula

    j = c/f (j=wavelength; f= frequency) which implies that c= j*f.

    using m as the mass of the electron= 9.11*10^-31kg, You can get your answer.

    Else,

    THERE IS NOTHING ELSE I CAN PROPOSE HERE…

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