Problems on Rutherford model of the atom (Serway et al. p.138-9)

Answers (R. Egerton)

8. (a) particles/minute = C [sin(phi/2)]^-4 where C = constant here

so 100 = C [sin(10deg)]^-4 and C = 100 [.1736]^4 = 0.0909

particles/min at 40 deg. = 6.64

particles/min at 60 deg. = 1.45

particles/min at 80 deg. = 0.264

(b) particles/minute is proportional to (speed)^-4 or (kinetic energy)^-2

So doubling the kinetic energy will result in 100/(2^2) = 25 particles/min

(c) If angle is constant, particle rate is proportional to (Z^2)N, where

N is the number of nuclei per unit area of the foil

= (atoms/volume)(foil thickness) = (mass/vol.)(mass/atom)(thickness)

=(mass/vol)(atomic weight)(atomic mass unit)(thickness)

So for same thickness of Cu, rate = 100 (29/79)^2 (197/63.5) (8.9/79.3)

= 19.3 particles/minute

9. For a head-on collision at the point of closest approach (distance r from center of nucleus), potential energy= (Ze)(2e) / [(4 pi epsilon)( r)]

= original kinetic energy = (13.9E6)(1.6E-19) J, giving r = 6.0E-15 m

Bohr atom (Serway et al. p.139):

12. Lyman series corresponds to nf = 1 in the formula: hf = R [(1/nf^2) – (1/ni^2)]

For the first Lyman line, ni = 2, giving hf = (13.6 eV) [1 – 0.25] = 10.2 eV = 1.63E-18 J

Wavelength = (hc/hf) = (1.986E-25)/(1.63E-18) = 1.21E-7 m

For the second line, ni = 3, giving hf = (13.6 eV) [1 – 0.1111] = 12.09 eV = 1.94E-18 J

Wavelength = (hc/hf) = (1.986E-25)/(1.94E-18) = 1.02E-7 m

For the third line, ni = 4, giving hf = (13.6 eV)[1 – 0.0625] = 12.75 eV = 2.04E-18 J

Wavelength = (hc/hf) = (1.986E-25)/(2.04E-18) = 9.72E-8 m

16. For the one-electron system Li(2+), En = (-13.6 eV) [Z^2 / n^2] = (-13.6eV)(9/n^2)

Ground state (n=1) hs an energy of –122.4 eV

Second energy level (n=2) has an energy of (-122.4)/4 = - 30.6 eV

Third level has an energy of (-122.4)/9 = -13.6 eV

22. In hydrogen ground state, total energy = -13.6 eV = -(ke^2)/(2r) - see Eq.(3.27)

Kinetic energy = +(ke^2)/(2r) = 13.6 eV

Potential energy = total energy – kinetic energy

= -(ke^2)/(2r) - (ke^2)/(2r) = - ke^2 / r = -27.2 eV