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weqwewe [10]
3 years ago
14

The first ionization energy, e, of a potassium atom is 0.696 aj. what is the wavelength of light, in nm, that is just sufficient

to ionize a potassium atom? values for constants can be found here.
Chemistry
1 answer:
elena55 [62]3 years ago
6 0
The formula to be used for this problem is as follows:

E = hc/λ, where h is the Planck's constant, c is the speed of light and λ is the wavelength. Also 1 aJ = 10⁻¹⁸ J

0.696×10⁻¹⁸ = (6.62607004×10⁻³⁴ m²·kg/s)(3×10⁸ m/s)/λ
Solving for λ,
λ = 2.656×10⁻⁷ m or <em>0.022656 nm</em>
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