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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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Answer:

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While alkynes have 3 bonds, it must be understood that they have 2 pi bonds only and as such their degree of saturation is two.

In the case of an alkene, there is only one single pi bond and as such the degree of unsaturation is 1.

Now in this case, we have a fractional 0.5 degree of unsaturation alongside the 3 to make a total of 3.5. So what’s the issue here?

The fractional part shows that the chemical entity we are dealing with here is a radical. While the integer 3 shows that there are 3 pi-bonds, the half pi bond remaining tells us that there is a missing electron on one of the atoms involved in the chemical bonding and as such, the 1/2 extra degree of unsaturation tends to tell us this.

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