(a) An electron moves with a speed of `4.70xx10^6` m//s. What is its de-Broglie wavelength?
(b) A proton moves with the same speed. Determine its de - Broglie wavelength.

Explanation

To solve the problem, we will calculate the de Broglie wavelength for both the electron and the proton moving at the same speed of \(4.70 \times 10^6\) m/s.

### Part (a): De Broglie Wavelength of the Electron

1. **Identify the formula for de Broglie wavelength**:
The de Broglie wavelength (\(\lambda\)) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant and \(p\) is the momentum.

2. **Express momentum in terms of mass and velocity**:
The momentum \(p\) can be expressed as:
\[ p = mv \]
where \(m\) is the mass of the particle and \(v\) is its velocity.

3. **Substitute the expression for momentum into the wavelength formula**:
Therefore, the de Broglie wavelength can be rewritten as:
\[ \lambda = \frac{h}{mv} \]

4. **Insert known values**:
- Planck's constant \(h = 6.626 \times 10^{-34} \, \text{J s}\)
- Mass of the electron \(m_e = 9.11 \times 10^{-31} \, \text{kg}\)
- Speed \(v = 4.70 \times 10^6 \, \text{m/s}\)

Now plug these values into the equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{(9.11 \times 10^{-31})(4.70 \times 10^6)} \]

5. **Calculate the denominator**:
\[ (9.11 \times 10^{-31})(4.70 \times 10^6) = 4.28 \times 10^{-24} \, \text{kg m/s} \]

6. **Calculate the wavelength**:
\[ \lambda = \frac{6.626 \times 10^{-34}}{4.28 \times 10^{-24}} \approx 1.55 \times 10^{-10} \, \text{m} \]

### Part (b): De Broglie Wavelength of the Proton

1. **Use the same formula for de Broglie wavelength**:
\[ \lambda = \frac{h}{mv} \]

2. **Insert known values for the proton**:
- Mass of the proton \(m_p = 1.67 \times 10^{-27} \, \text{kg}\)
- Speed \(v = 4.70 \times 10^6 \, \text{m/s}\)

Now plug these values into the equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{(1.67 \times 10^{-27})(4.70 \times 10^6)} \]

3. **Calculate the denominator**:
\[ (1.67 \times 10^{-27})(4.70 \times 10^6) = 7.85 \times 10^{-21} \, \text{kg m/s} \]

4. **Calculate the wavelength**:
\[ \lambda = \frac{6.626 \times 10^{-34}}{7.85 \times 10^{-21}} \approx 8.44 \times 10^{-14} \, \text{m} \]

### Final Answers:
(a) The de Broglie wavelength of the electron is approximately \(1.55 \times 10^{-10} \, \text{m}\).

(b) The de Broglie wavelength of the proton is approximately \(8.44 \times 10^{-14} \, \text{m}\).