Which of the following is true concerning the electropositive atom Al? Aluminum

Question

Which of the following is true concerning the electropositive atom Al? Aluminum will lose one electron to form its most stable ion Al and the radius of Al^+ is less than the radius of Al. Aluminum will lose three electrons to become Al^3 + and the radius of Al^3 + is the less as the radius of Al. Aluminum will lose three electrons to become Al^3 - and the radius of A^3 - is greater than the radius of Al. None of these. Which of the following is true concerning the following four species: 48_Ti, 41_K^+, 40_Ar, 31_P^3 - The order of increasing radius is 48_Ti

Explanation / Answer

Q10.

Apply Rydberg Formula

E = R*(1/nf^2 – 1/ni ^2)

R = -2.178*10^-18 J

Nf = final stage/level

Ni = initial stage/level

E = Energy per unit (i.e. J/photon)

E = (-2.178*10^-18)*(1/3^2 – 1/6 ^2)

E = 1.815*10^-19 J/particle

For the wavelength:

WL = h c / E

h = Planck Constant = 6.626*10^-34 J s

c = speed of particle (i.e. light) = 3*10^8 m/s

E = energy per particle J/photon

WL = (6.626*10^-34)(3*10^8)/(1.815*10^-19)

WL = 0.0000010952 m

to nanometers:

WL = (0.0000010952)(10^9) = 1095.2 nm

answer is A

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