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Chapter 28: Problem 9

What is the ratio of the wavelength of a 0.100-keV photon to the wavelength of a 0.100-keV electron?

Short Answer

Expert verified
The ratio of the wavelength of a 0.100-keV photon to the wavelength of a 0.100-keV electron is approximately 32:1.

Step by step solution

01

Planck's equation for photons

We will use Planck's equation to find the wavelength of the photon. Planck's equation relates the energy \(E_p\) of a photon to its wavelength \(\lambda_p\) and is given by: $$ E_p = h \frac{c}{\lambda_p} $$ where \(h\) is the Planck's constant (\(6.626 \times 10^{-34} Js\)) and \(c\) is the speed of light (\(3 \times 10^8 m/s\)).
02

de Broglie wavelength equation for electrons

For the electron, we'll use the de Broglie wavelength equation, which relates the energy \(E_e\) of an electron to its wavelength \(\lambda_e\). The equation is given by: $$ \lambda_e = \frac{h}{\sqrt{2 m_e E_e}} $$ where \(m_e\) is the mass of the electron (\(9.109 \times 10^{-31} kg\)).
03

Convert energy to Joules

Both energy values given are in keV, we need to convert them to Joules for further calculations. The conversion factor is \(1 keV = 1.602 \times 10^{-16} J\). Thus, $$ E_p = E_e = 0.1\text{ keV} \times 1.602 \times 10^{-16}J/keV = 1.602 \times 10^{-17}J $$
04

Calculate the photon's wavelength

Now, we can calculate the wavelength of the photon using Planck's equation: $$ \lambda_p = \frac{hc}{E_p} = \frac{6.626 \times 10^{-34} Js \times 3 \times 10^{8} m/s}{1.602 \times 10^{-17}J} = 1.239 \times 10^{-9} m $$
05

Calculate the electron's wavelength

Next, we calculate the wavelength of the electron using de Broglie wavelength equation: $$ \lambda_e = \frac{h}{\sqrt{2 m_e E_e}} = \frac{6.626 \times 10^{-34} Js}{\sqrt{2 \times 9.109 \times 10^{-31} kg \times 1.602 \times 10^{-17} J}} = 3.870 \times 10^{-11} m $$
06

Find the ratio of the two wavelengths

Finally, we will find the ratio of the wavelength of the photon to the wavelength of the electron: $$ \frac{\lambda_p}{\lambda_e} = \frac{1.239 \times 10^{-9} m}{3.870 \times 10^{-11} m} \approx 32.03 $$ The ratio of the wavelength of a 0.100-keV photon to the wavelength of a 0.100-keV electron is approximately 32:1.

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Most popular questions from this chapter

How many electron states of the H atom have the quantum numbers \(n=3\) and \(\ell=1 ?\) Identify each state by listing its quantum numbers.
An electron is confined in a one-dimensional box of length \(L\). Another electron is confined in a box of length 2 \(L\). Both are in the ground state. What is the ratio of their energies $E_{2 l} / E_{L} ?$
An electron is confined to a one-dimensional box of length \(L\). When the electron makes a transition from its first excited state to the ground state, it emits a photon of energy 0.20 eV. (a) What is the ground-state energy (in electron-volts) of the electron in this box? (b) What are the energies (in electron-volts) of the photons that can be emitted when the electron starts in its third excited state and makes transitions downwards to the ground state (either directly or through intervening states)? (c) Sketch the wave function of the electron in the third excited state. (d) If the box were somehow made longer, how would the electron's new energy level spacings compare with its old ones? (Would they be greater, smaller, or the same? Or is more information needed to answer this question? Explain.)
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