Let's tackle each part of your question step by step.

1. Classical Uncertainty | Response time - bandwidth relation (2 pt)

(a) Maximum time-span of a single pulse

The rate of pulses is given as 100,000 pulses/s. The time period (T) for one pulse can be calculated as the inverse of the pulse rate:

$T = \frac{1}{\text{Rate}} = \frac{1}{100,000 \, \text{pulses/s}} = 10^{-5} \, \text{s} = 10 \, \mu s$

This means the maximum time-span for a single pulse, such that subsequent pulses do not overlap, is 10 microseconds.

(b) Range of pulse frequencies (f_min, f_max)

The frequency of the pulses corresponds directly to their rate. The minimum frequency (f_min) is essentially the frequency at which a single pulse occurs, while the maximum frequency (f_max) would correspond to the highest rate that can still be recognized by the receiving equipment.

Given:

• $f = \text{Rate} = 100,000 \, \text{pulses/s}$

Thus,

• $f_{min} = 100,000 \, \text{Hz}$
• $f_{max}$ would be determined by the bandwidth limitations of the receiver but can be approximated by the Nyquist rate, which is typically twice the maximum frequency for accurate signal representation.

For simplicity, without additional details on bandwidth, we can consider $f_{max}$ to be around twice the pulse frequency:

• $f_{max} \approx 200,000 \, \text{Hz}$

2. Uncertainty Principle (E,t) (1 pt)

The uncertainty in energy ($\Delta E$) can be calculated using the uncertainty principle:

$\Delta E \cdot \Delta t \geq \frac{\hbar}{2}$

Given that the lifetime of the state ($\Delta t$) is $10^{-9} \, \text{s}$ and $\hbar \approx 10^{-34} \, \text{J·s}$:

$\Delta E \geq \frac{\hbar}{2 \Delta t} = \frac{10^{-34} \, \text{J·s}}{2 \times 10^{-9} \, \text{s}} = 5 \times 10^{-26} \, \text{J}$

Thus, the uncertainty in the energy of photons emitted by the atoms is approximately $5 \times 10^{-26} \, \text{J}$.

3. Position-Momentum Uncertainty Principle (2 pt)

Using the position-momentum uncertainty principle:

$\Delta x \cdot \Delta p \geq \frac{\hbar}{2}$

Where:

• $\Delta x$ (uncertainty in position) = $10^{-2} \, \text{mm} = 10^{-5} \, \text{m}$
• The mass of the beetle (m) = $1.0 \, \text{mg} = 1.0 \times 10^{-6} \, \text{kg}$

The momentum ($p$) is related to velocity ($v$) by:

$p = mv$

Thus:

$\Delta p = m \Delta v$

Substituting into the uncertainty principle:

$10^{-5} \cdot m \Delta v \geq \frac{10^{-34}}{2}$

Calculating $m \Delta v$:

$\Delta v \geq \frac{10^{-34}}{2 \cdot 10^{-5} \cdot 10^{-6}} = \frac{10^{-34}}{10^{-11}} = 10^{-23} \, \text{m/s}$

Thus, the beetle might be moving at a speed uncertainty of approximately $10^{-23} \, \text{m/s}$.

Summary of Results:

1. (a) Maximum time-span of a pulse: 10 µs (b) Pulse frequency range: $f_{min} = 100,000 \, \text{Hz}, f_{max} \approx 200,000 \, \text{Hz}$
2. Uncertainty in energy: $5 \times 10^{-26} \, \text{J}$
3. Speed uncertainty for the beetle: $10^{-23} \, \text{m/s}$

Would you like any further details or clarifications? Here are some questions to consider:

1. How does the bandwidth of a receiver affect its ability to pick up signals?
2. Can you explain more about the implications of the uncertainty principle in quantum mechanics?
3. What practical applications exist for measuring uncertainty in energy?
4. How does the uncertainty in position relate to the observable behavior of particles?
5. What are some other scenarios where these principles might apply?

Tip: When dealing with uncertainties, always ensure units are consistent throughout calculations to avoid errors.