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Chapter 7: Problem 56

Calculate the velocities of electrons with de Broglie wavelengths of \(1.0 \times 10^{2} \mathrm{~nm}\) and \(1.0 \mathrm{~nm}\), respectively.

Short Answer

Expert verified
The velocities of electrons with de Broglie wavelengths of \(1.0 \times 10^2 \mathrm{~nm}\) and \(1.0 \mathrm{~nm}\) are \(7.27 \times 10^5 \mathrm{\dfrac{m}{s}}\) and \(7.27 \times 10^7 \mathrm{\dfrac{m}{s}}\) respectively.

Step by step solution

01

Rewrite the de Broglie equation in terms of velocity

We need to find the velocity, so we need to rewrite the given formula in terms of velocity. We have the momentum \(p = m \cdot v\). We can rewrite the de Broglie equation as: \( \lambda = \dfrac{h}{m \cdot v} \) Now, we can solve for the velocity: \(v = \dfrac{h}{\lambda \cdot m} \) We will use this equation to calculate the velocities.
02

Identify the constants and convert the wavelengths to meters

In the equation we derived, we have: - \(h\) (Planck's constant) = \(6.626 \times 10^{-34} \mathrm{Js}\) - \(m\) (mass of an electron) = \(9.109 \times 10^{-31} \mathrm{kg}\) Our given wavelengths are in nanometers. We need to convert them to meters to use them in the equation. \(1.0 \times 10^2 \mathrm{~nm} = 1.0 \times 10^{2} \times 10^{-9} \mathrm{m} = 1.0 \times 10^{-7} \mathrm{m}\) \(1.0 \mathrm{~nm} = 1.0 \times 10^{-9} \mathrm{m}\) We'll now use these values and the equation to calculate the velocities.
03

Calculate the velocities for de Broglie wavelengths of \(1.0 \times 10^2 \mathrm{~nm}\) and \(1.0 \mathrm{~nm}\)

For the de Broglie wavelength of \(1.0 \times 10^2 \mathrm{~nm}\): \(v_{1} = \dfrac{6.626 \times 10^{-34} \mathrm{Js}}{(1.0 \times 10^{-7} \mathrm{m})(9.109 \times 10^{-31} \mathrm{kg})} \) \(v_{1} = 7.27 \times 10^5 \mathrm{\dfrac{m}{s}}\) For the de Broglie wavelength of \(1.0 \mathrm{~nm}\): \(v_{2} = \dfrac{6.626 \times 10^{-34} \mathrm{Js}}{(1.0 \times 10^{-9} \mathrm{m})(9.109 \times 10^{-31} \mathrm{kg})} \) \(v_{2} = 7.27 \times 10^7 \mathrm{\dfrac{m}{s}}\) So, the velocities of electrons with de Broglie wavelengths of \(1.0 \times 10^2 \mathrm{~nm}\) and \(1.0 \mathrm{~nm}\) are \(7.27 \times 10^5 \mathrm{\dfrac{m}{s}}\) and \(7.27 \times 10^7 \mathrm{\dfrac{m}{s}}\) respectively.

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Most popular questions from this chapter

How many unpaired electrons are present in each of the following in the ground state: \(\mathrm{O}, \mathrm{O}^{+}, \mathrm{O}^{-}, \mathrm{Os}, \mathrm{Zr}, \mathrm{S}, \mathrm{F}\), Ar?

Which has the more negative electron affinity, the oxygen atom or the \(\mathrm{O}^{-}\) ion? Explain your answer.

Does a photon of visible light \((\lambda \approx 400\) to \(700 \mathrm{~nm})\) have sufficient energy to excite an electron in a hydrogen atom from the \(n=1\) to the \(n=5\) energy state? from the \(n=2\) to the \(n=6\) energy state?

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