The de Broglie wavelength of a ___ will have the shortest wavelength when traveling at $$ 30 \mathrm{~cm} / \mathrm{s} $$. hydrogen atom uranium atom planet Submit Request Answer

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The de Broglie wavelength is given by

$$
\lambda = \frac{h}{mv},
$$

where  
$$h$$ is Planck’s constant,  
$$m$$ is the mass of the object,  
$$v$$ is the velocity.

For a fixed velocity $$v = 30 \text{ cm/s}$$, the wavelength $$\lambda$$ is inversely proportional to the mass $$m$$. This means that the larger the mass, the smaller (shorter) the de Broglie wavelength.

Let's compare the mass of the options:

- **Hydrogen atom:** very light, with a mass of approximately $$1\,\text{amu} \approx 1.67 \times 10^{-27} \, \text{kg}$$.
- **Uranium atom:** much heavier than a hydrogen atom, roughly $$238\,\text{amu} \approx 3.95 \times 10^{-25} \, \text{kg}$$.
- **Planet:** an astronomical object with an extremely large mass, many orders of magnitude larger than a uranium atom.

Since the de Broglie wavelength decreases with increasing mass, the planet will have a much shorter wavelength compared to the atoms when both are traveling at $$30 \text{ cm/s}$$.

Thus, the de Broglie wavelength will be shortest for a planet.
A planet will have the shortest de Broglie wavelength when traveling at $$30 \text{ cm/s}$$.

The de Broglie wavelength of a ___ will have the shortest wavelength when traveling at \( 30 \mathrm{~cm} / \mathrm{s} \). hydrogen atom uranium atom planet Submit Request Answer

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