Find the de-Broglie wavelengths of (a) a 46 g golf ball with a velocity of 30m/s (b) an electron with a velocity of `10^7 m//s.`

Explanation

To find the de-Broglie wavelengths for the given cases, we will use the de-Broglie wavelength formula:

\[ \lambda = \frac{h}{p} = \frac{h}{mv} \]

where:
- \(\lambda\) is the de-Broglie wavelength,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(m\) is the mass of the object,
- \(v\) is the velocity of the object.

### (a) For the golf ball:

1. **Convert mass from grams to kilograms:**
\[ m = 46 \, \text{g} = 46 \times 10^{-3} \, \text{kg} = 0.046 \, \text{kg} \]

2. **Use the given velocity:**
\[ v = 30 \, \text{m/s} \]

3. **Substitute the values into the de-Broglie wavelength formula:**
\[ \lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{(0.046)(30)} \]

4. **Calculate the denominator:**
\[ mv = 0.046 \times 30 = 1.38 \, \text{kg m/s} \]

5. **Calculate the de-Broglie wavelength:**
\[ \lambda = \frac{6.626 \times 10^{-34}}{1.38} \approx 4.8 \times 10^{-34} \, \text{m} \]

### (b) For the electron:

1. **Use the mass of the electron:**
\[ m = 9.1 \times 10^{-31} \, \text{kg} \]

2. **Use the given velocity:**
\[ v = 10^7 \, \text{m/s} \]

3. **Substitute the values into the de-Broglie wavelength formula:**
\[ \lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31})(10^7)} \]

4. **Calculate the denominator:**
\[ mv = 9.1 \times 10^{-31} \times 10^7 = 9.1 \times 10^{-24} \, \text{kg m/s} \]

5. **Calculate the de-Broglie wavelength:**
\[ \lambda = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-24}} \approx 7.28 \times 10^{-11} \, \text{m} \]

### Summary of Results:
- (a) The de-Broglie wavelength of the golf ball is approximately \(4.8 \times 10^{-34} \, \text{m}\).
- (b) The de-Broglie wavelength of the electron is approximately \(7.28 \times 10^{-11} \, \text{m}\).