What minimum frequency of light is required to ionize potassium in419 KJ/mol?
1.05 x 10^15
To determine the minimum frequency of light required to ionize potassium, we need to use the equation for calculating the energy of a photon:
E = hf
Where E is the energy of the photon, h is Planck's constant (6.626 × 10^-34 Joule seconds), and f is the frequency of the light.
The minimum energy required to ionize potassium can be obtained by converting 419 kJ/mol to Joules per atom:
Energy per atom = (419 kJ/mol) / (6.022 × 10^23 atoms/mol)
Once we have the energy per atom, we can divide it by the Avogadro constant to get the energy per photon:
Energy per photon = Energy per atom / Avogadro constant
Finally, we rearrange the equation to solve for the frequency:
f = E / h
Let's do the calculations:
Energy per atom = (419 kJ/mol) / (6.022 × 10^23 atoms/mol)
Energy per photon = Energy per atom / Avogadro constant
Frequency of light = Energy per photon / Planck's constant
Plugging in the values:
Energy per atom = (419 × 10^3 J) / (6.022 × 10^23 atoms/mol) ≈ 6.963 × 10^-20 J
Energy per photon = 6.963 × 10^-20 J / (6.022 × 10^23) ≈ 1.157 × 10^-43 J
Frequency of light = 1.157 × 10^-43 J / (6.626 × 10^-34 J·s) ≈ 1.747 × 10^10 Hz
Therefore, the minimum frequency of light required to ionize potassium is approximately 1.747 × 10^10 Hz.