# Solutions 13

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## S13.1

Use the relationship between momentum and velocity and Heisenberg's uncertainty principle to derive a formula for the uncertainty of velocity.

$$p = m_e v \]$$ \Delta p = m_e \Delta v \]

$$\Delta x \Delta p \geq \frac{h}{4 \pi} \]$$ \Delta v \geq \frac{h}{4 \pi m_e \Delta x} \]

Plug in the given values

$$\Delta v \geq \frac{6.626 \times 10^{-34}\ J\ s}{4 \pi \times 9.109 \times 10^{-31}\ kg \times 2.0\ Å \times \frac{m}{10^{10}\ Å}} \times \frac{\frac{kg\ m^2}{s^2}}{J} \]$$ \Delta v \geq 2.9 \times 10^5\ m/s \]

## S13.2

Use the direct application of the uncertainty principle:

$\Delta x \Delta p \geq \dfrac{h}{4\pi}$

Let's consider the molecule has an uncertainly that is ±radius±radius of the balloon.

Δx=1.25×10−5m

The uncertainty of the momentum of the molecule can be estimated via the uncertainty principle.

$\Delta p = \dfrac{h}{4\pi\Delta x}=\dfrac{6.626 \times 10^{-34} J s}{4\pi \times 1.25\times10^{-5}m} \times \frac{\frac{kg\ m^2}{s^2}}{J} =4.22 \times 10^{-30} kg m s^{-1}$

This can be converted to uncertainty in velocity via

p=mv

or

Δp=mΔv

The mass of the oxygen molecule is:

$m=\dfrac{M_{O_{2}} \times 1 kg/1000 g}{N_{A}}=\dfrac{32 g/mol \times 1 kg/1000g}{6.02 \times10^{23} mol^{-1}}=5.32\times10^{-26} kg$

Therefore,

$Δv=\dfrac{Δp}{m}=\dfrac{4.22 \times 10^{-30} kg m s^{-1}}{5.32 \times 10^{-26} kg}=7.93 \times 10^{-5} m s^{-1}$

## S13.3

B, C, D cannot have physical significance.

Reasons:

B: The wavefunction has discontinuity;

C: The value of the wavefunction goes to infinity as horizontal axis approaches zero (will make the probability infinity which is not physicallly possible);

D: The wavefunction does not have single value.

## S13.4

The tunneling probability can be expressed as

$P=exp[-\dfrac{4 \pi a}{h} [2m(V-E)]^{1/2}]$

where a is the width of the barrier a=7 x 10-10 m.

The mass of the electron is 9.1x10-31 kg and V = 6 eV x 1.6*10-19 J/eV = 9.6 x 10-19 J.

Take the natural log of the equation:

$ln(P)=-\dfrac{4 \pi a}{h}{2m(V-E)}^{1/2}=-2 a [\dfrac{2m/}{ħ^{2}}(V-E)]^{1/2}$

So,

$[\dfrac{2m}{ħ^{2}}(V-E)]^{1/2}=-\dfrac{0.001}{2 \times 7 \times 10^{-10} m}=4.93 \times 10^9 m^{-1}$

$ħ=\dfrac{h}{2\pi}$

$E=V-\dfrac{ħ ^{2} (4.93 \times 10^9 m^{-1})^{2}}{2m}=9.6 \times 10^{-19} J - \dfrac{(1.054 \times 10^{-34} J s \times 4.93 \times 10^{9} m^{-1})^2}{2\times 9.1 \times 10^{-31} kg}=8.11 \times 10^{-19} J = 5.07 eV$

An electron with E=5.07 eV has a probability of 0.001 or 0.1% of tunneling through the 6 eV barrier.

## S13.5

Identify the similarities and differences between the time-independent Schrödinger equations and the equation governing the law of conservation of energy.

Similarities: Both equations are saying that the sum of kinetic energy and potential energy is the total energy;

Differences: In the equation governing the law of conservation of enegy, T and V are kinetic and potential energy functions, whereas in the time-independent Schrodinger equation, the T and V functions are replaced with kinetic energy and potential energy operators.

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