Barron's Big Idea 1 Calculations

Barron's Big Idea 1

Calculations

Q8

l = c/v = 3.00 x 10^8/4.00 x 10^14 = 0.750 x 10-6 m = 750 nm

Choose C

Free-response

(b)

350 000/6.02 x 10^23 = 5.81 x 10^-19 J/bond

E = hv thus v = E/h = 5.81 x 10^-19/6.63 x 10^-34 = 8.76 x 10^14 s-1

c = lv then l = c/v = 3.00 x 10^8/8.76 x 10^14 = 3.42 x 10^-7 m = 342 nm

thus if the photon has a wavelength of 342 nm it is able to break the bond.

and 342 nm is at the upper end of the ultraviolet rigion.

any light with a shorter wavelength is ok to ionize (break) the bond.

(d)

first use Rydberg’s equation:

(-2.178 x 10-18 J) /n^2 = E

thus at n = 2

E = (-2.178 x 10-18 J)/4 = -0.5445 x 10^-18 J

at n = 5

E = (-2.178 x 10-18 J)/25 = -0.08712 x 10^-18 J

so the energy gap is

(-0.08712 + 0.5445) x 10^-18 = 4.57 x 10^-19 J

then E = hv, v = 4.57 x 10^-19/6.63 x 10^-34 = 6.89 x 10^14 s-1

c = lv, then l = c/v = 3.00 x 10^8/6.89 x 10^14 = 0.435 x 10^-6, thus 435 nm

435 nm is visible light