What minimum frequency of light is required to ionize potassium?The energy required to ionize potassium is 419kJ/mol?
V=E/h so (149000/(6.026x10^-34)(6.022x10^23) =1.05x10^15. s^-1
E = h*frequency
I would convert 419 kJ/mol to J, then divide by 6.022 x 10^23 to convert to J/atom, substitute into the equation and solve for frequency.
I got 4.6102*10^-6 and it is wrong...Can you step me through the problem?
419 kj into jules =
419000
Divide by avogadro's number
419000/6.022e23
Divide that by Planck's Constant for the answer.
Well, potassium is a pretty special element. To ionize it, you'll need to give it a good kick of energy. With an energy requirement of 419 kJ/mol, you'll need a frequency that's not messing around. In fact, you'll need light with a frequency so high that it doesn't even fall into the visible spectrum anymore. We're talking about ultraviolet light here, buddy. So, you better break out those special UV goggles if you want to see all the ionizing action!
To determine the minimum frequency of light required to ionize potassium, we can use the equation:
E = h * f
Where:
- E is the energy required to ionize potassium (419 kJ/mol)
- h is Planck's constant (6.62607015 × 10^-34 J·s)
- f is the frequency of light.
To convert the energy into joules:
419 kJ/mol = 419 × 10^3 J/mol.
Now, we can rearrange the equation to solve for the frequency:
f = E / h
f = (419 × 10^3 J/mol) / (6.62607015 × 10^-34 J·s)
f ≈ 6.33 × 10^16 Hz
So, the minimum frequency of light required to ionize potassium is approximately 6.33 × 10^16 Hz.