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Ch 38: Quantization
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Capitolo 38, Problema 68a

INT A beam of electrons is incident upon a gas of hydrogen atoms. What minimum speed must the electrons have to cause the emission of 656 nm light from the 3→2 transition of hydrogen?

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Determine the energy of the emitted photon corresponding to the 656 nm wavelength using the equation for photon energy: \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the emitted light.
Calculate the energy difference between the \( n=3 \) and \( n=2 \) energy levels of the hydrogen atom using the formula for the energy levels of hydrogen: \( E_n = -\frac{13.6}{n^2} \; \text{eV} \). Subtract \( E_2 \) from \( E_3 \) to find the energy required for the transition.
Recognize that for the 3→2 transition to occur, the incident electron must transfer at least this energy to the hydrogen atom. This means the kinetic energy of the electron must be equal to or greater than the energy difference calculated in the previous step.
Relate the kinetic energy of the electron to its speed using the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the electron and \( v \) is its speed. Solve for \( v \) by setting \( KE \) equal to the energy difference from step 2.
Substitute the known values for the mass of the electron, the energy difference, and any necessary constants into the equation to calculate the minimum speed of the electron.

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Photon Emission and Energy Levels

In hydrogen atoms, electrons occupy discrete energy levels. When an electron transitions from a higher energy level (n=3) to a lower one (n=2), it emits a photon with a specific wavelength, in this case, 656 nm. The energy of the emitted photon corresponds to the difference in energy between these two levels, which can be calculated using the formula E = hc/λ, where h is Planck's constant and c is the speed of light.
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Kinetic Energy of Electrons

The kinetic energy of an electron is given by the equation KE = 1/2 mv², where m is the mass of the electron and v is its velocity. To cause the emission of a photon, the incident electron must have sufficient kinetic energy to excite the hydrogen atom to the required energy level. This means the kinetic energy must be equal to or greater than the energy difference between the initial and final states of the hydrogen atom.
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Threshold Energy and Ionization

Threshold energy refers to the minimum energy required to initiate a process, such as exciting an electron in an atom. For the hydrogen atom's 3→2 transition, the incident electron must have enough energy to overcome the energy difference between these levels. If the energy is too low, no photon will be emitted; if it is too high, it may lead to ionization, where the electron is completely removed from the atom.
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