Certainly! Let's solve the problem step-by-step.
### Step-by-Step Solution
1. Understand the De Broglie Wavelength Formula:
The de Broglie wavelength ([tex]\(\lambda\)[/tex]) can be calculated using the formula:
[tex]\[
\lambda = \frac{h}{mv}
\][/tex]
where:
- [tex]\(h\)[/tex] is Planck's constant ([tex]\(6.63 \times 10^{-34} \, J \cdot s\)[/tex])
- [tex]\(m\)[/tex] is the mass of the object (in kg)
- [tex]\(v\)[/tex] is the velocity of the object (in m/s)
2. Given Data:
- Mass ([tex]\(m\)[/tex]) = 0.030 kg
- Velocity ([tex]\(v\)[/tex]) = 540 m/s
- Planck's Constant ([tex]\(h\)[/tex]) = [tex]\(6.63 \times 10^{-34} \, J \cdot s\)[/tex]
3. Substitute the Values into the Formula:
[tex]\[
\lambda = \frac{6.63 \times 10^{-34}}{0.030 \times 540}
\][/tex]
4. Perform the Calculation:
First, calculate the denominator:
[tex]\[
0.030 \times 540 = 16.2
\][/tex]
Then divide Planck's constant by this product:
[tex]\[
\lambda = \frac{6.63 \times 10^{-34}}{16.2}
\approx 4.092592592592593 \times 10^{-35} \, \text{m}
\][/tex]
5. Round to Significant Figures:
The answer [tex]\(4.092592592592593 \times 10^{-35} \, \text{m}\)[/tex] can be rounded to match the significant figures typically provided in such contexts. Hence, it is approximately [tex]\(4.1 \times 10^{-35} \, \text{m}\)[/tex].
### Conclusion
Hence, the de Broglie wavelength of the bullet of mass 0.030 kg traveling at [tex]\(540 \, \text{m/s}\)[/tex] is approximately [tex]\(4.1 \times 10^{-35} \, \text{m}\)[/tex].
The correct answer from the given options is:
[tex]\[ \boxed{4.1 \times 10^{-35} \, \text{m}} \][/tex]
