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How fast must a \(56.5-g\) tennis ball travel to have a de Broglie wavelength equal to that of a photon of green light ( \(5400 \mathrm{~A}\) )?

Short Answer

Expert verified
The tennis ball must travel at approximately \(2.16 \times 10^{-27} \mathrm{~m/s}\).

Step by step solution

01

Understand the de Broglie wavelength formula

The de Broglie wavelength is given by the formula \(\lambda = \frac{h}{p}\), where \(\lambda\) is the wavelength, \(h\) is Planck's constant \(6.626 \times 10^{-34} \mathrm{~J} \, \mathrm{s}\), and \(p\) is the momentum of the particle. Momentum \(p\) can be written as \(p = mv\), where \(m\) is the mass and \(v\) is the velocity.
02

Convert quantities to proper units

Convert the mass of the tennis ball from grams to kilograms: \(56.5 \mathrm{~g} = 0.0565 \mathrm{~kg}\). Also, convert the wavelength from \(\mathrm{~A}\) (Angstroms) to meters: \(5400 \mathrm{~A} = 5400 \times 10^{-10} \mathrm{~m} = 5.4 \times 10^{-7} \mathrm{~m}\).
03

Rearrange the de Broglie equation

Since we need to find the velocity \(v\), rearrange the de Broglie equation to solve for \(v\): \(\lambda = \frac{h}{mv} \Rightarrow v = \frac{h}{m\lambda}\).
04

Plug in the values

Insert the known values into the equation: \(v = \frac{6.626 \times 10^{-34} \mathrm{~J} \, \mathrm{s}}{(0.0565 \mathrm{~kg}) (5.4 \times 10^{-7} \mathrm{~m})}\).
05

Calculate the velocity

Perform the calculation: \(v = \frac{6.626 \times 10^{-34}}{0.0565 \times 5.4 \times 10^{-7}} \approx 2.16 \times 10^{-27} \mathrm{~m/s}\).
06

Interpret the result

The velocity obtained is extremely small, on the order of \(10^{-27} \mathrm{~m/s}\), indicating that the tennis ball would have to move incredibly slowly to have a de Broglie wavelength equal to that of a photon of green light.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Planck's constant
Planck's constant is a fundamental constant in quantum mechanics that relates the energy of a photon to its frequency. It is a key element in the equation for the de Broglie wavelength. Planck's constant is denoted as \(h\) and is equal to \(6.626 \times 10^{-34} \text{ J} \times \text{s}\). This constant is crucial for calculating the wavelength of both particles and waves, bridging the gap between classical and quantum physics. By understanding Planck's constant, we can explore the wave-particle duality of matter, which is fundamental to quantum mechanics.
Momentum equation
Momentum is a measure of the motion of an object and is given by the product of its mass and velocity. For any object, the momentum \(p\) can be calculated using the equation \( p = mv \), where \(m\) is the object's mass and \(v\) is its velocity. In the context of the de Broglie wavelength, momentum helps bridge the gap between a particle's motion and its wave characteristics. By combining the momentum equation with the de Broglie formula \( \lambda = \frac{h}{p} \), we can derive the wavelength of a particle based on its mass and velocity.

Understanding this allows us to see that even objects with mass, like a tennis ball, exhibit wave-like properties under certain conditions.
Wavelength conversion
To solve problems involving wavelengths, it’s important to convert units to standard measurements—typically meters in scientific contexts. In the example, we converted the green light wavelength from angstroms to meters. Since \( 1 \text{ \AA} = 1 \times 10^{-10} \text{ m} \), the original 5400 \text{ \AA} becomes \( 5400 \times 10^{-10} \text{ m} = 5.4 \times 10^{-7} \text{ m} \). This conversion is essential for plugging into the de Broglie equation.

Similarly, converting the mass of the tennis ball from grams to kilograms ensures all quantities are in compatible units, simplifying calculations and reducing the chance of errors.
Mass and velocity relationship
The relationship between mass and velocity is crucial in understanding how we calculate the de Broglie wavelength. From the de Broglie equation \( \lambda = \frac{h}{mv} \), we see that the wavelength depends inversely on both mass and velocity. This means:
  • Larger mass or higher velocity results in a shorter wavelength.
  • Smaller mass or lower velocity yields a longer wavelength.
When solving for velocity, as in our tennis ball example, we rearrange the equation to \( v = \frac{h}{m\lambda} \). By inputting the known values of Planck's constant, mass, and wavelength, we can solve for the velocity, reflecting how these quantities are interdependent in determining the wave properties of matter.

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Most popular questions from this chapter

The flame tests for sodium and potassium are based on the emissions at \(589 \mathrm{nm}\) and \(404 \mathrm{nm},\) respectively. When both elements are present, the \(\mathrm{Na}^{+}\) emission is so strong that the \(\mathrm{K}^{+}\) emission can be seen only by looking through a cobalt-glass filter. (a) What are the colors of the \(\mathrm{Na}^{+}\) and \(\mathrm{K}^{+}\) emissions? (b) What does the cobalt-glass filter do? (c) Why is \(\mathrm{KClO}_{4}\) used as an oxidizing agent in fireworks rather than \(\mathrm{NaClO}_{4} ?\)

The quantum-mechanical treatment of the hydrogen atom gives this expression for the wave function, \(\psi,\) of the \(1 s\) orbital: $$ \Psi=\frac{1}{\sqrt{\pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2} e^{-r l a_{0}} $$ where \(r\) is the distance from the nucleus and \(a_{0}\) is \(52.92 \mathrm{pm}\). The probability of finding the electron in a tiny volume at distance \(r\) from the nucleus is proportional to \(\psi^{2}\). The total probability of finding the electron at all points at distance \(r\) from the nucleus is proportional to \(4 \pi r^{2} \psi^{2}\). Calculate the values (to three significant figures) of \(\psi, \psi^{2},\) and \(4 \pi r^{2} \psi^{2}\) to fill in the following table and sketch a plot of each set of values versus \(r .\) Compare the latter two plots with those in Figure \(7.17 \mathrm{~A}\).

The quantum-mechanical treatment of the \(\mathrm{H}\) atom gives the energy, \(E\), of the electron as a function of \(n:\) $$E=-\frac{h^{2}}{8 \pi^{2} m_{e} a_{0}^{2} n^{2}} \quad(n=1,2,3, \ldots) $$where \(h\) is Planck's constant, \(m_{e}\) is the electron's mass, and \(a_{0}\) is \(52.92 \times 10^{-12} \mathrm{~m}\) (a) Write the expression in the form \(E=-\) (constant) \(\left(1 / n^{2}\right),\) evaluate the constant (in \(J\) ), and compare it with the corresponding expression from Bohr's theory. (b) Use the expression from part (a) to find \(\Delta E\) between \(n=2\) and \(n=3\) (c) Calculate the wavelength of the photon that corresponds to this energy change.

In addition to continuous radiation, fluorescent lamps emit some visible lines from mercury. A prominent line has a wavelength of \(436 \mathrm{nm}\). What is the energy (in \(\mathrm{J}\) ) of one photon of it?

Explain the significance of the fact that the peak in the radial probability distribution plot for the \(n=1\) level of an \(\mathrm{H}\) atom is at 0.529 A. Is the probability of finding an electron at 0.529 A from the nucleus greater for the \(1 s\) or the \(2 s\) orbital?

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