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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.

Respuesta :

meerkat18
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 0.022656 nm

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