Consider the one-dimensional potential step defined by
U(x) = 0 , x < 0, U(x) = U_{0, }x > 0.

Suppose a wave incident from the left has energy E = 4U_{0}.
What is the probability that the wave will be reflected?

Solution:

- Concepts:

This is a "square potential" problem. We solve HΦ(x) = EΦ(x) in regions where U(x) is constant and apply boundary conditions. - Reasoning:

We are given a piecewise constant potential. We have a potential that has a value of 0 for x < 0 and a value of U_{0}(barrier) for x > 0. We are asked to find the the reflectance R. - Details of the calculation:

We have E > U_{0}.

The most general solutions in regions 1, and 2 are

Φ_{1}(x) = A_{1}exp(ik_{1}x) + A_{1}'exp(-ik_{1}x),

Φ_{2}(x) = A_{2}exp(ik_{2}x),

for particles incident from the left.

Here k_{1}^{2}= (2m/ħ^{2})E and k_{2}^{2}= (2m/ħ^{2})(E - U_{0}).

Φ is continuous at x = 0. Therefore A_{1 }+ A_{1}' = A_{2}.

(∂/∂x)Φ(x) is continuous at x = 0. Therefore ik_{1}A_{1 }- ik_{1}A_{1}' = ik_{2}A_{2}.

We can then solve for A_{1}'/A_{1}. We find A_{1}'/A_{1}= (k_{1}- k_{2})/(k_{1}+ k_{2}).

The**reflectance**R = |A_{1}'/A_{1}|^{2}_{ }= 1 - 4k1k_{2}/(k_{1}+ k_{2})^{2}.

k_{1}^{2}= 8mU_{0}/ħ^{2}, k_{2}^{2}= 6mU_{0}/ħ^{2}, k_{1}k_{2}= (48)^{½}(mU_{0}/ħ^{2}), R = 1 - 4(48)^{½}/(√8 + √6)^{2}= 0.005

Consider a one-dimensional step potential, of the form

U(x) = 0 for x < 0,

U(x) = U_{0} for x > 0, with U_{0} > 0.

A particle with mass m and energy E > U_{0} is incident on this step
from the left.

(a) Write down the appropriate solutions for the time-independent Schroedinger equation for this particle in the x < 0 region and in the x > 0
region.

(b) Apply the appropriate boundary conditions at x = 0 to match these
solutions.

(c) Derive expressions for the probabilities that the particle is reflected
and transmitted by the step.

Solution:

- Concepts:

Square potentials, the potential step - Reasoning:

We are given a piecewise constant potential and are asked to find the probability of reflection (R) and transmission (T ). - Details of the calculation:

(a) Write the time-independent Schroedinger equation as

(∂^{2}/∂x^{2})Φ(x) + k^{2}Φ(x) = 0. E - U = ħ^{2}k^{2}/(2m).

Let k_{1}= (2mE/ħ^{2})^{½}, k_{2}= (2m(E – U_{0})/ħ^{2})^{½},

Then in region 1 (x < 0) Φ_{1}(x) = A_{1}exp(ik_{1}x) + A_{1}'exp(-ik_{1}x)

and in region 2 (x > 0) Φ_{2}(x) = A_{2}exp(ik_{2}x) + A_{1}'exp(-ik_{2}x)

are the most general solutions.

(b) Φ is continuous at x = 0. Therefore A_{1 }+ A_{1}' = A_{2 }+A_{2}' .

∂Φ/∂x is continuous at x = 0. Therefore ik_{1}A_{1 }- ik_{1}A_{1}' = ik_{2}A_{2 }- ik_{2}A_{2}' .

We have two equations and four unknowns. No unique solution exists. Let us limit ourselves to particles approaching from the left being transmitted or reflected at the barrier. Then A_{2}' = 0. We can then solve for A_{1}'/A_{1}and A_{2}/A_{1}. We find

A_{1}'/A_{1}= (k_{1}– k_{2})/(k_{1}+ k_{2}) and A_{2}/A_{1}= 2k_{1}/(k_{1}+ k_{2})

The ratios are real and positive, there is no phase shift upon reflection and transmission.

(c) Our solutions in regions 1 and 2 are plane wave solutions. They are not square integrable, and a proper normalization is not possible. Plane wave solutions are useful in describing steady streams of particles. To evaluate the probability of reflection and transmission for such beams, we compare the reflected and transmitted flux with the incident flux. Let P(x) be the probability density at x, let x_{0}be some point in region 1, and let x_{0}^{'}be some point in region 2. We then have

incident flux ∝ v_{1}P_{i}(x_{0}), reflected flux ∝ v_{1}P_{r}(x_{0}), transmitted flux ∝ v_{2}P_{t}(x_{0}').

The reflectance is R = v_{1}P_{r}(x_{0})/(v_{1}P_{i}(x_{0})) = |A_{1}'/A_{1}|^{2}= (k_{1}– k_{2})^{2}/(k_{1}+ k_{2})^{2},

and the transmittance is T = v_{2}P_{t}(x_{0}^{'})/(v_{1}P_{i}(x_{0})) = (k_{2}/k_{1})|A_{1}'/A_{1}|^{2}= (4k_{2}k_{1})/(k_{1}+ k_{2})^{2}.

We have R = 1 - (4k_{2}k_{1})/(k_{1}+ k_{2})^{2}, T = (4k_{2}k_{1})/(k_{1}+ k_{2})^{2}, R + T = 1.

Consider a one-dimensional quantum-mechanical scattering problem, involving a
particle of mass m moving through a region with potential energy function

U(x) = U_{0}, 0 ≤ x ≤ L, U(x) = 0 otherwise.

The particle moves from -∞ to +∞. Assume that its energy is chosen to be
exactly U_{0}. Find the transmission and the
reflection probabilities.

Solution:

- Concepts:

This is a "square potential" problem. We solve HΦ(x) = EΦ(x) in regions where U(x) is constant and apply boundary conditions. - Reasoning:

Assume U(x) = U = constant in certain regions of space. In such a region the Schroedinger equation yields (∂^{2}/∂x^{2})Φ(x) + (2m/ħ^{2})(E - U)Φ(x) = 0.

Let E > U: (∂^{2}/∂x^{2})Φ(x) + k^{2}Φ(x) = 0. E - U = ħ^{2}k^{2}/(2m).

The most general solution is Φ(x) = Aexp(ikx) + A'exp(-ikx), with A and A' complex constants.

Let E < U: (∂^{2}/∂x^{2})Φ(x) - ρ^{2}Φ(x) = 0. U - E = h^{2}ρ^{2}/(2m).

The most general solution is Φ(x) = Bexp(ρx) + B'exp(-ρx), with B and B' complex constants.

Let E = U: (∂^{2}/∂x^{2})Φ(x) = 0. Φ(x) = Cx + C', with C and C' complex constants. - Details of the calculation:

At a finite step the boundary conditions are that Φ(x) and (∂/∂x)Φ(x) are continuous.

Let x < 0, define region 1, 0 < x < L define region 2, and x > L define region 3, and let

E = U_{0}= h^{2}k^{2}/(2m).

The most general solutions in regions 1, 2, and 3 are:

Φ_{1}(x) = A_{1}exp(ikx) + A_{1}'exp(-ikx), Φ_{2}(x) = A_{2}x + A_{2}', Φ_{3}(x) = A_{3}exp(ikx)

Boundary conditions:

x = 0: (1) A_{1}+ A_{1}' = A_{2}', (2) ikA_{1 }- ikA_{1}' = A_{2}

x = L: (3) A_{2}L + A_{2}'= A_{3}exp(ikL), (4) A_{2}= ikA_{3}exp(ikL).

R = |A_{1}'/A_{1}|^{2}, T = |A_{3}/A_{1}|^{2}.

From (1) and (2): A_{1}+ A_{1}' = A_{2}', A_{1 }- A_{1}' = A_{2}/(ik),

A_{1}= ½(A_{2}' + A_{2}/(ik)), A_{1}' = ½(A_{2}' - A_{2}/(ik)).

A_{1}'/A_{1}= (A_{2}'/A_{2}- 1/(ik))/(A_{2}'/A_{2}+ 1/(ik))

From (3) and (4): A_{2}L + A_{2}'= A_{2}/(ik), A_{2}'/A_{2}= 1/(ik) – L.

A_{1}'/A_{1}= -L/(2/(ik) - L) = Lk/(2i + Lk).

R = (Lk)^{2}/(4 + (Lk)^{2}). T = 1 - R = 4/(4 + (Lk)^{2}).

Let U(x) = 0 for x < 0, U(x) = U_{0 }> 0 for 0 < x < a, and
U(x) = 0 for x > a. A particle of mass m and energy E > U_{0} is incident on this potential barrier. Find the
energies E for which resonances occur, i.e. for which the probability of
transmission is 1.

Solution:

- Concepts:

This is a "square potential" problem. We solve HΦ(x) = EΦ(x) in regions where U(x) is constant and apply boundary conditions. - Reasoning:

We are given a piecewise constant potential and are asked to find the transmission coefficient T and the condition for T = 1. - Details of the calculation:

The most general solutions in regions 1, 2, and 3 are

Φ_{1}(x) = A_{1}exp(ik_{1}x) + A_{1}'exp(-ik_{1}x),

Φ_{2}(x) = A_{2}exp(ik_{2}x) + A_{2}'exp(-ik_{2}x),

Φ_{3}(x) = A_{3}exp(ik_{1}x).

Here k_{1}^{2}= (2m/ħ^{2})E and k_{2}^{2}= (2m/ħ^{2})(E - U_{0}).

The boundary conditions are that Φ(x) and (∂/∂x)Φ(x) are continuous at x = 0 and x = a.

x = 0: A_{1}+ A_{1}' = A_{2 }+ A_{2}' , ik_{1}A_{1 }- ik_{1}A_{1}' = ik_{2}A_{2 }- ik_{2}A_{2}'.

x = a: A_{2}exp(ik_{2}a) + A_{2}'exp(-ik_{2}a) = A_{3}exp(ik_{1}a),

ik_{2}A_{2}exp(ik_{2}a) - ik_{2}A_{2}'exp(-ik_{2}a) = ik_{1}A_{3}exp(ik_{1}a).

We need to solve these equations for A_{3}in terms of A_{1}.

(i) Solve for A_{2}and A_{2}' in terms of A_{3}.

A_{2}= ½exp(i(k_{1}-k_{2})a)(1 + k_{1}/k_{2})A_{3}= C A_{3}.

A_{2}' = ½exp(i(k_{1}+k_{2})a)(1 - k_{1}/k_{2})A_{3}= C' A_{3}.

(ii) Now solve for A_{1}in terms of A_{3}.

2A_{1}= A_{2}+ A_{2}' + (k_{2}/k_{1})(A_{2 }- A_{2}') = (C + C' + (k_{2}/k_{1})(C - C'))A_{3}.

A_{1}= ([-i(k_{1}^{2}+ k_{2}^{2}))/(2k_{1}k_{2})]sink_{2}a + cosk_{2}a)exp(ik_{1}a)A_{3}.

T = |A_{3}/A_{1}|^{2}= 4k_{1}^{2}k_{2}^{2}/[(k_{1}^{2}+ k_{2}^{2})^{2}sin^{2}k_{2}a + 4k_{1}^{2}k_{2}^{2}cos^{2}k_{2}a]

= 4k_{1}^{2}k_{2}^{2}/[(k_{1}^{2}- k_{2}^{2})^{2}sin^{2}k_{2}a + 4k_{1}^{2}k_{2}^{2}].

T = 1 is sin^{2}k_{2}a = 0, k_{2}a = nπ, k_{2}^{2}= (2m/ħ^{2})(E - U_{0}) = n^{2}π^{2}/a^{2},

E = (ħ^{2}/(2m))n^{2}π^{2}/a^{2}+ U_{0}, n = 1, 2, ... .

A plane wave representing an electron beam with energy E is
incident from the negative x direction onto a potential energy step described by
the function

U(x) = U_{2} for x < 0,

U(x) = U_{1} for 0 < x < a,

U(x) = 0 for x > a,

where U_{1} = π^{2}ħ^{2}/(8ma^{2}),
E = 2U_{1}, U_{1} < U_{2} < E.

Evaluate the transmittance T. For which value of U_{2}
is T the largest?

Solution:

- Concepts:

Square potentials - Reasoning:

We are given a piecewise constant potential and are asked to find the transmission coefficient T. - Details of the calculation:

The most general solutions in regions 1, 2, and 3 are

Φ_{1}(x) = A_{1}exp(ik_{1}x) + A_{1}'exp(-ik_{1}x),

Φ_{2}(x) = A_{2}exp(ik_{2}x) + A_{2}'exp(-ik_{2}x),

Φ_{3}(x) = A_{3}exp(ik_{3}x).

Here k_{1}^{2}= (2m/ħ^{2})(2U_{1}– U_{2}) and k_{2}^{2}= (2m/ħ^{2})U_{1}, k_{3}^{2}= (2m/ħ^{2})2U_{1}.

The boundary conditions are that Φ(x) and (∂/∂x)Φ(x) are continuous at x = 0 and x = a.

x = 0: A_{1}+ A_{1}' = A_{2 }+ A_{2}' , ik_{1}A_{1 }- ik_{1}A_{1}' = ik_{2}A_{2 }- ik_{2}A_{2}'.

x = a: A_{2}exp(ik_{2}a) + A_{2}'exp(-ik_{2}a) = A_{3}exp(ik_{3}a),

ik_{2}A_{2}exp(ik_{2}a) - ik_{2}A_{2}'exp(-ik_{2}a) = ik_{3}A_{3}exp(ik_{3}a).

We need to solve these equations for A_{3}in terms of A_{1}.

(i) Solve for A_{2}and A_{2}' in terms of A_{3}.

A_{2}= ½exp(i(k_{3}-k_{2})a)(1 + k_{3}/k_{2})A_{3}= C A_{3}.

A_{2}' = ½exp(i(k_{3}+k_{2})a)(1 – k_{3}/k_{2})A_{3}= C' A_{3}.(ii) Now solve for A

_{1}in terms of A_{3}.

2A_{1}= A_{2}+ A_{2}' + (k_{2}/k_{1})(A_{2 }- A_{2}') = (C + C' + (k_{2}/k_{1})(C - C'))A_{3}.

C + C' = ½exp(i(k_{3}a)[(2cos(k_{2}a) - (k_{3}/k_{2})2i sin(k_{2}a)]

C - C' = ½exp(i(k_{3}a)[-2i sin(k_{2}a) + (k_{3}/k_{2}) 2cos(k_{2}a)]

C + C' + (k_{2}/k_{1})(C - C') = exp(i(k_{3}a)[(1 + k_{3}/k_{1}) cos(k_{2}a) – i((k_{3}/k_{2}+ k_{2}/k_{1}) sin(k_{2}a)]

A_{1}= {[-i(k_{1}k_{3}+ k_{2}^{2}))/(2k_{1}k_{2})]sink_{2}a + [(k_{1}k_{2}+ k_{3}k_{2})/(2k_{1}k_{2})]cosk_{2}a}exp(ik_{1}a)A_{3}.

|A_{3}/A_{1}|^{2}= 4k_{1}^{2}k_{2}^{2}/[(k_{1}k_{3}+ k_{2}^{2})^{2}sin^{2}k_{2}a + (k_{1}k_{2}+ k_{3}k_{2})^{2}(cos^{2}k_{2}a)]

In this problem since k_{2}^{2}a^{2}= (2m/ħ^{2})U_{1}a^{2}= π^{2}/4, sin^{2}k_{2}a = 1, cos^{2}k_{2}a = 0, k_{2}^{2}= ½k_{3}^{2}.

|A_{3}/A_{1}|^{2}= 4k_{1}^{2}k_{2}^{2}/(k_{1}k_{3}+ k_{2}^{2})^{2}= 2k_{1}^{2}k_{3}^{2}/(k_{1}k_{3}+ ½k_{3}^{2})^{2}= 8/(2 + k_{3}/k_{1})^{2}.

T = (k_{3}/k_{1})| |A_{3}/A_{1}|^{2}= 8(k_{3}/k_{1}) /(2 + k_{3}/k_{1})^{2}.

Let X = k_{3}/k_{1}, X^{2}= 2U_{1}/(2U_{1}– U_{2}), 2 < X^{2}< ∞, since U_{1}< U_{2}< 2U_{1}.

T = 8X/(2 + X)^{2},

dT/dX = 0 --> 8(2 + X) - 16X = 0, X = 2, X^{2}= 2U_{1}/(2U_{1}– U_{2}) = 4, U_{2}= 3U_{½}.

T is largest for U_{2}= 3U_{½}. Then T = 1.

(a)
Calculate the transmission coefficient for a particle with mass m and kinetic
energy E passing through the rectangular potential barrier

U{x) = 0 for x < 0, U(x) = U_{0} for 0 < x < a, U(x) = 0 for x
> a, with E < U_{0}.

(b) Show that for E << U_{0} and 2mU_{0}a^{2}/ħ^{2}
>> 1 the transmittance T can be written as

T ~ (16E/U_{0})exp[-2(2m(U_{0}
a^{2})/ħ^{2})^{½}].

Solution:

- Concepts:

This is a "square potential" problem. We solve HΦ(x) = EΦ(x) in regions where U(x) is constant and apply boundary conditions. - Reasoning:

We are given a piecewise constant potential. We are asked to find the transmission coefficient t. - Details of the calculation:

Divide space into 3 regions; region 1: x' < 0, region 2: 0 < x' < a, region 3: x' > a.

(a) potential barrier, E < U_{0}.

The most general solutions in regions 1, 2, and 3 are

Φ_{1}(x') = A_{1}exp(ik_{1}x') + A_{1}'exp(-ik_{1}x'),

Φ_{2}(x') = A_{2}exp(ρ_{2}x') + A_{2}'exp(-ρ_{2}x'),

Φ_{3}(x') = A_{3}exp(ik_{1}x').

Here k_{1}^{2}= (2m/ħ^{2})E and ρ_{2}^{2}= (2m/ħ^{2})(U_{0}- E).

The boundary conditions are that Φ(x') and (d/dx')Φ(x') are continuous at x' = 0 and x' = a.

x = 0: A_{1}+ A_{1}' = A_{2 }+ A_{2}' , ik_{1}A_{1 }- ik_{1}A_{1}' = ρ_{2}A_{2 }- ρ_{2}A_{2}'.

x = a: A_{2}exp(ρ_{2}a) + A_{2}'exp(-ρ_{2}) = A_{3}exp(ik_{1}a),

ρ_{2}A_{2}exp(ik_{2}a) - ρ_{2}A_{2}'exp(-ρ_{2}) = ik_{1}A_{3}exp(ik_{1}a).

We need to solve these equations for A_{3}in terms of A_{1}.

(i) Solve for A_{2}and A_{2}' in terms of A_{3}.

A_{2}= ½ exp((ik_{1}- ρ_{ 2})a)(1 + ik_{1}/ρ_{ 2})A_{3}= C A_{3}.

A_{2}' = ½ exp((ik_{1}+ ρ_{ 2})a)(1 - ik_{1}/ρ_{ 2})A_{3}= C' A_{3}.

(ii) Now solve for A_{1}in terms of A_{3}.

2A_{1}= A_{2}+ A_{2}' + (ρ_{2}/ik_{1})(A_{2 }- A_{2}') = (C + C' + (ρ_{2}/ik_{1})(C - C'))A_{3}.

A_{1}= [[(k_{1}^{2}- ρ_{2}^{2})/(2ik_{1}ρ_{2})]sinhρ_{2}a + coshρ_{2}a ] exp(ik_{1}a)A_{3}.

t = A_{3}/A_{1}= exp(-ik_{1}a)/[[(k_{1}^{2}- ρ_{2}^{2})/(2ik_{1}ρ_{2})]sinhρ_{2}a + coshρ_{2}a]

T = |A_{3}/A_{1}|^{2}= 4k_{1}^{2}ρ_{2}^{2}/[(k_{1}^{2}- ρ_{2}^{2})^{2}sinh^{2}ρ_{2}a + 4k_{1}^{2}ρ_{2}^{2}cosh^{2}ρ_{2}a]

= 4k_{1}^{2}ρ_{2}^{2}/[(k_{1}^{2}+ ρ_{2}^{2})^{2}sinh^{2}ρ_{2}a + 4k_{1}^{2}ρ_{2}^{2}].

[cosh^{2}x - sinh^{2}x = 1.]

T = 4E(U_{0}- E) / [(U_{0}^{2}sinh^{2}[(2m(U_{0}- E)/ħ^{2})^{½}a] + 4E(U_{0}- E)].

(b) Let E << U_{0}and 2mU_{0}a^{2}/ħ^{2}>> 1.

sinh(x) = ½ (e^{x}– e^{-x}). If x >> 1 sinh(x) ~ ½ e^{x}.

T ~ 4EU_{0}/[(U_{0}^{2}/4)exp[2(2m(U_{0}a^{2})/ħ^{2})^{½}] + 4EU_{0}] = 1/ [(U_{0}/(16E))exp[2(2m(U_{0}a^{2})/ħ^{2})^{½}+ 1]

T ~ (16E/U_{0})exp[-2(2m(U_{0}a^{2})/ħ^{2})^{½}].

(a) Calculate the transmission coefficient for a particle with
mass m and kinetic energy E < U_{0} passing through the rectangular potential
barrier

U(x) = 0 for x < 0, U(x) = U_{0 }> 0 for 0 < x < a, and
U(x) = 0 for x > a.

(b) Show that for E << U_{0} and 2mU_{0}a^{2}/ħ^{2}
>> 1 the transmission coefficient can be written as
T ≈ (16E/U_{0})exp[-2(2mU_{0}/ħ^{2})^{½}a].

(c) Many heavy nuclei decay by emitting an alpha particle.
In a
simple one-dimensional model, the potential barrier the alpha particles have to penetrate
can be approximated by

U(r) = 0 for r < R_{0}, U(r)
= U_{0}R_{0}/r for r > R_{0},

where R_{0} is the radius of the nucleus and
U_{0} is the
barrier height for r_{0} = R. The energy E of the alpha particle can
be assumed to be much smaller than U_{0}. For a non constant potential
barrier the expression for the transmission coefficient found in part (b) can be used as a
guide. Assume that for E << U_{0}, we have T ≈ exp[-2∫_{R1}^{R2 }dr (2m(U(r) - E)/ħ^{2})^{½}].

The integration limits R_{1}
and R_{2} are determined as solutions to the equation U(r) = E.

Calculate the alpha transmission coefficient and the decay constant λ,
i.e. the decay probability per second.

Solution:

- Concepts:

This is a "square potential" problem. We solve HΦ(x) = EΦ(x) in regions where U(x) is constant and apply boundary conditions. - Reasoning:

We are given a piecewise constant potential. We are asked to find the transmittance T. - Details of the calculation:

(a) Divide space into 3 regions;

region 1: x < 0, region 2: 0 < x < a, region 3: x > a.

(a) potential barrier, E < U_{0}.

The most general solutions in regions 1, 2, and 3 are

Φ_{1}(x) = A_{1}exp(ikx) + A_{1}'exp(-ikx),

Φ_{2}(x) = B_{2}exp(ρx) + B_{2}'exp(-ρx),

Φ_{3}(x) = A_{3}exp(ikx).

Here k^{2}= (2m/ħ^{2})E and ρ^{2}= (2m/ħ^{2})(U_{0}- E).

The boundary conditions are that Φ(x) and (∂/∂x)Φ(x) are continuous at x = 0 and x = a.

x = 0: A_{1}+ A_{1}' = B_{2 }+ B_{2}', ikA_{1 }- ikA_{1}' = ρB_{2 }- ρB_{2}'.

x = a: B_{2}exp(ρa) + B_{2}'exp(-ρa) = A_{3}exp(ika),

ρB_{2}exp(ik_{2}a) - ρB_{2}'exp(-ρa) = ik_{i}A_{3}exp(ika).

We need to solve these equations for A_{3}and A_{1}' in terms of A_{1}.

(i) Solve for B_{2}and B_{2}' in terms of A_{3}.

B_{2}= ½exp((ik - ρ)a)(1 + ik/ρ )A_{3}= C A_{3}.

B_{2}' = ½exp((ik + ρ)a)(1 - ik/ρ)A_{3}= C' A_{3}.

(ii) Now solve for A_{1}in terms of A_{3}to find T.

2A_{1}= B_{2}+ B_{2}' + (ρ/ik)(B_{2 }- B_{2}') = (C + C' + (ρ/ik) (C - C'))A_{3}.

A_{1}= ([(k^{2}- ρ^{2})/(2ikρ)]sinhρa + coshρa)exp(ika)A_{3}.

T = |A_{3}/A_{1}|^{2}= 4k^{2}ρ^{2}/[(k^{2}- ρ^{2})^{2}sinh^{2}ρa + 4k^{2}ρ^{2}cosh^{2}ρa]

= 4k^{2}ρ^{2}/[(k^{2}+ ρ^{2})^{2}sinh^{2}ρa + 4k^{2}ρ^{2}].

[cosh^{2}x - sinh^{2}x = 1.]

T = 4E(U_{0}- E)/[U_{0}^{2}sinh^{2}[(2m(U_{0}- E)/ħ^{2})^{½}a] + 4E(U_{0}- E)].(b) E << U

_{0}and 2mU_{0}a^{2}/ħ^{2}>> 1.

sinh(x) = ½(e^{x}- e^{-x}) ≈ ½e^{x}, if x >> 1.

Therefore

T ≈ 4E(U_{0}- E)/[¼U_{0}^{2}exp[2(2m(U_{0}- E)/ħ^{2})^{½}a] + 4E(U_{0}- E)]

≈ 4EU_{0}/[¼U_{0}^{2}exp[2(2mU_{0}/ħ^{2})^{½}a] + 4EU_{0})]

≈ (16E/U_{0})exp[-2(2mU_{0}/ħ^{2})^{½}a].(c) Assume that T ≈ exp[-2∫

_{R1}^{R2 }dr (2m(U(r) - E)/ħ^{2})^{½}].

R

_{1}= R_{0}, R_{2}= U_{0}R_{0}/E, R_{2}>> R_{1}.

∫_{R1}^{R2 }dr (U(r) - E)^{½}= ∫_{R1}^{R2 }dr (U_{0}R_{0}/r - U_{0}R_{0}/R_{2})^{½}

= (U_{0}R_{0})^{½}∫_{R1}^{R2 }dr (1/r - 1/R_{2})^{½}.

∫_{R1}^{R2 }dr (1/r - 1/R_{2})^{½}= ∫_{R1}^{R2 }dr [(R_{2}- r)/(rR_{2})^{]½}

= √R_{2}[(R_{1}/R_{2}- (R_{1}/R_{2})^{2})^{½}+ π/2 - tan^{-1}(R_{1}/(R_{2}- R_{1}))^{½}

--> √R_{2}π/2 if R_{2}>> R_{1}.

T ≈ exp[-√(2m/E) πU_{0}R_{0}/ħ) = e^{-λ'}, λ' = √(2m/E) πU_{0}R_{0}/ħ

e^{-λ'}is the transmission coefficient.

To calculate the escape probability per second, we have to multiply T by the rate of the alpha particle hitting the barrier, which is approximately v/R_{0}.

escape probability ≈ √(2m/E)/R_{0}* e^{-λ'}= λ = 1/τ.

This is a reasonable order of magnitude approximation as long as the escape probability is very small.