A proton `(m_(p) = 1.673 xx 10^(-27) kg)` and an electron `(m = 9.109 xx 10^(-31) kg)` are confined such that the position x of each is known within `1.50 xx 10^(-10) m`. What is the ratio of the minimum uncertainty in the x component of the velocity of the electron to that of the proton, `Deltav_(e)// Deltav_(p)`?

Answer

To find the ratio of the minimum uncertainty in the x-component of the velocity of the electron to that of the proton, we can use the Heisenberg Uncertainty Principle. The principle states that:

\[ \Delta x \Delta p \geq \frac{h}{4\pi} \]

where:
- \(\Delta x\) is the uncertainty in position,
- \(\Delta p\) is the uncertainty in momentum,
- \(h\) is Planck's constant (\(h \approx 6.626 \times 10^{-34} \, \text{Js}\)).

The momentum \(p\) of a particle is given by:

\[ p = mv \]

where \(m\) is the mass and \(v\) is the velocity. Therefore, the uncertainty in momentum can be expressed as:

\[ \Delta p = m \Delta v \]

Substituting this into the uncertainty principle gives:

\[ \Delta x \cdot (m \Delta v) \geq \frac{h}{4\pi} \]

From this, we can solve for \(\Delta v\):

\[ \Delta v \geq \frac{h}{4\pi m \Delta x} \]

Now, we will calculate \(\Delta v\) for both the electron and the proton.

1. **Calculate \(\Delta v_e\) for the electron:**
- Mass of the electron \(m_e = 9.109 \times 10^{-31} \, \text{kg}\)
- Uncertainty in position \(\Delta x = 1.50 \times 10^{-10} \, \text{m}\)

\[ \Delta v_e \geq \frac{h}{4\pi m_e \Delta x} \]

Substituting the values:

\[ \Delta v_e \geq \frac{6.626 \times 10^{-34}}{4\pi (9.109 \times 10^{-31})(1.50 \times 10^{-10})} \]

Calculating this gives:

\[ \Delta v_e \geq \frac{6.626 \times 10^{-34}}{4\pi \times 9.109 \times 10^{-31} \times 1.50 \times 10^{-10}} \]

Evaluating the denominator:

\[ \approx 4\pi \times 9.109 \times 10^{-31} \times 1.50 \times 10^{-10} \approx 8.577 \times 10^{-40} \]

So,

\[ \Delta v_e \geq \frac{6.626 \times 10^{-34}}{8.577 \times 10^{-40}} \approx 7.73 \times 10^{5} \, \text{m/s} \]

2. **Calculate \(\Delta v_p\) for the proton:**
- Mass of the proton \(m_p = 1.673 \times 10^{-27} \, \text{kg}\)

\[ \Delta v_p \geq \frac{h}{4\pi m_p \Delta x} \]

Substituting the values:

\[ \Delta v_p \geq \frac{6.626 \times 10^{-34}}{4\pi (1.673 \times 10^{-27})(1.50 \times 10^{-10})} \]

Evaluating the denominator:

\[ \approx 4\pi \times 1.673 \times 10^{-27} \times 1.50 \times 10^{-10} \approx 3.152 \times 10^{-36} \]

So,

\[ \Delta v_p \geq \frac{6.626 \times 10^{-34}}{3.152 \times 10^{-36}} \approx 2.10 \times 10^{2} \, \text{m/s} \]

3. **Calculate the ratio \(\frac{\Delta v_e}{\Delta v_p}\):**

\[ \frac{\Delta v_e}{\Delta v_p} = \frac{7.73 \times 10^{5}}{2.10 \times 10^{2}} \approx 3,683.33 \]

Thus, the ratio of the minimum uncertainty in the x-component of the velocity of the electron to that of the proton is approximately:

\[ \frac{\Delta v_e}{\Delta v_p} \approx 3,683 \]