__Bohr atom__

Consider a Bohr-type, classical model of a hydrogenic atom, where a
structureless electron with charge -q_{e} is orbiting a (stationary)
structureless nucleus with charge Zq_{e} in a circular orbit in the xy-plane
without radiating. Let the angular momentum vector of the electron be **L**
= L**k** and have magnitude L = nħ.

(a) What is the orbital radius r_{0} of the electron?

(b) What is the kinetic energy of the electron and what is its total energy?

(c) What is the magnetic dipole moment **μ** of the atom?

Solution:

- Concepts:

The classical Bohr atom - Reasoning:

The Bohr atom introduces angular momentum quantization, but otherwise is a classical model. - Details of the calculation:

(a) kZq_{e}^{2}/r_{0}^{2}= mv^{2}/r_{0}, mvr_{0}= nħ, r_{0}= n^{2}ħ^{2}/(mkZq_{e}^{2}). Let α = kZq_{e}^{2}. r_{0}= L^{2}/(mα).

(b) ½mv^{2}= ½α/r_{0}= mα^{ 2}/(2n^{2}ħ^{2}). E_{total}= T + U = -½α/r_{0}= -m(kZq_{e}^{2})^{2}/(2n^{2}ħ^{2}).

(c)**μ**= IA**n**= -(q_{e}v/(2πr))(πr^{2})**n**= = -½q_{e}vr = -(q_{e}/(2m))**L**= -μ_{B}n**k**.

__Nuclei and particles__

Calculate the binding energy of the deuteron, which consists of a proton and a neutron, given that the mass of a deuteron is 2.013553 u.

Solution:

- Concepts:

Nuclear binding energy - Reasoning:

The binding energy represents how much energy we would have to supply to pull the nucleus apart into separate free nucleons. - Details of the calculation:

Binding energy: (m_{p}+ m_{n})c^{2}– m_{d}c^{2}= 2.2 MeV

Consider the case of a chain of two radioactive decays. Nucleus
A decays into another
B by one process, then
B decays into another
C by a second process, i.e.
A → B → C.

The mean life of A is τ_{A} = 1/λ_{A}, and the mean life of B is
τ_{B} = 1/λ_{B},

Initially N_{0A} nuclei of type A are present and N_{0B} = 0.

(a) Find the decay rate (the number of decays per second) for B in terms of N_{0A},
λ_{A}, and λ_{B},

(b) Find approximate expressions for the decay rate if

(i) λ_{A} >> λ_{B}, (ii) λ_{A} << λ_{B},
(iii) λ_{A} ≈ λ_{B},

Solution:

- Concepts:

Radioactive decay - Reasoning:

The decay of an unstable nucleus is a random (probabilistic) process. The rate at which nuclei decay is proportional to the number N of nuclei present. - Details of the calculation:

(a) dN_{A}/dt = - λ_{A}N_{A}, N_{A}(t) = N_{0A}exp(-λ_{A}t).

dN_{B}/dt = λ_{A}N_{A}– λ_{B}N_{B}. (feed rate plus decay rate)

dN_{B}/dt = λ_{A}N_{0A}exp(-λ_{A}t) – λ_{B}N_{B}.

Try a solution N_{B}= C exp(-λ_{A}t) + D exp(-λ_{B}t).

D = - C since N_{B}(0) = 0.

N_{B}= C(exp(-λ_{A}t) - exp(-λ_{B}t)),

dN_{B}/dt = -C(λ_{A}exp(-λ_{A}t) - λ_{B}exp(-λ_{B}t)) = λ_{A}N_{0A}exp(-λ_{A}t) – λ_{B}C(exp(-λ_{A}t) - exp(-λ_{B}t)),

-Cλ_{A}= λ_{A}N_{0A}– λ_{B}C, C = λ_{A}N_{0A}/(λ_{B}- λ_{A}).

N_{B}(t) = [λ_{A}N_{0A}/(λ_{B}- λ_{A})](exp(-λ_{A}t) - exp(-λ_{B}t)).

Decay rate: R_{B}= –λ_{B}N_{B}= -[λ_{B}λ_{A}N_{0A}/(λ_{B}- λ_{A})](exp(-λ_{A}t) - exp(-λ_{B}t)).

(b) (i) If λ_{A}>> λ_{B}, R_{B}(t) ≈ –λ_{B}N_{0A}exp(-λ_{B}t). The mean life of A is very short compared to the mean live of B, and nearly all A nuclei have decayed into B nuclei in a time interval very much smaller than τ_{B}.

(ii) λ_{A}<< λ_{B}, R_{B}(t) ≈ -λ_{A}N_{0A}exp(-λ_{A}t). The mean life of A is very long compared to the mean live of B. When a nucleus A decays, the nucleus B that was formed is likely to decay in a time interval very much smaller than τ_{A}.

(iii) λ_{A}≈ λ_{B}, write λ_{A}= λ, λ_{B}≈ λ + ε.

R_{B}≈ -[λ^{2}N_{0A}/ε](exp(-λt) (1 – 1 + εt) = -λ^{2}N_{0A}t(exp(-λt).

As ε --> 0 this becomes the exact solution for λ_{A}= λ_{B}= λ.