Is this really the end? I was just getting into it....

Problem 1.
a + b)


Don't worry about the fact that the signs are different from what we discussed in class today - your answers to the other parts are still consistent with the solutions below, and will be correct.
c) Highest energy is the 6 eV transition.
d) lowest is the 1 eV transition.
e) This is the reverse of c and d - longest wavelength will be the 1 eV transition since E = hc/lambda
f) shortest wavelength is with the 6 eV transition, which has the highest energy.
g) The difference in energy between these two energy levels is 5 eV. This translates into a photon of wavelength 248 nanometers.
h) Three are possible.
i) The electron has KE = 7.12 x 10^-19 J = 4.45 eV. This is enough energy to cause an n = 1 to n = 3 transition, but not n = 1 to n = 4.

Problem 2.

1996B5.
a) Z = 102

b) K = 1/2*mv^2 = 8.42 MeV = 1.35 x 10^-12 J
alpha particle mass = (4 u)*(1.66 x 10^-27kg
v = (2*K/m)^1/2 = 2.02 x 10⁷ m/s

c) A correct explanation includes a reference to:
1. a change in mass/mass defect being converted into KE fo the alpha particle

2. conservation of energy - binding energy of the nucleus is converted into KE of the alpha particle during the decay.

3. work-energy theorem - the electric force does work on the alpha particle as it moves away from the nucleus, thereby changing its KE.