Ch.6 - Electronic Structure of Atoms

Brown - Chemistry: The Central Science 15th Edition

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Brown 15th Edition

Ch.6 - Electronic Structure of Atoms

Problem 33c

Chapter 6, Problem 33c

Molybdenum metal must absorb radiation with a minimum frequency of 1.09 * 1015 s - 1 before it can eject an electron from its surface via the photoelectric effect. (c) If molybdenum is irradiated with light of wavelength of 120 nm, what is the maximum possible kinetic energy of the emitted electrons?

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Identify the given values: the minimum frequency (\( \nu_0 = 1.09 \times 10^{15} \text{ s}^{-1} \)) and the wavelength of the incident light (\( \lambda = 120 \text{ nm} \)).

Convert the wavelength from nanometers to meters by using the conversion factor \(1 \text{ nm} = 1 \times 10^{-9} \text{ m}\).

Calculate the frequency (\( \nu \)) of the incident light using the speed of light equation: \( c = \lambda \nu \), where \( c = 3.00 \times 10^8 \text{ m/s} \).

Determine the energy of the incident photons using Planck's equation: \( E = h \nu \), where \( h = 6.626 \times 10^{-34} \text{ J s} \).

Calculate the maximum kinetic energy of the emitted electrons using the photoelectric equation: \( KE_{\text{max}} = E - h \nu_0 \), where \( h \nu_0 \) is the work function of molybdenum.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Photoelectric Effect

The photoelectric effect is the phenomenon where electrons are emitted from a material when it absorbs light or electromagnetic radiation. This effect demonstrates the particle nature of light, where photons must have sufficient energy to overcome the work function of the material to eject electrons. The energy of the incoming photons is directly related to their frequency, as described by the equation E = hf, where E is energy, h is Planck's constant, and f is frequency.

Energy-Frequency Relationship

The energy of a photon is directly proportional to its frequency, as given by the equation E = hf. In this equation, h is Planck's constant (approximately 6.626 x 10^-34 J·s). This relationship is crucial for determining whether the energy of incoming photons is sufficient to eject electrons from a material, as seen in the context of the photoelectric effect.

Kinetic Energy of Emitted Electrons

The kinetic energy (KE) of emitted electrons in the photoelectric effect can be calculated using the equation KE = E - W, where E is the energy of the incoming photon and W is the work function of the material. The work function is the minimum energy required to remove an electron from the surface of the material. If the energy of the incoming photon exceeds the work function, the excess energy is converted into the kinetic energy of the emitted electron.

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Related Practice

Textbook Question

Classify each of the following statements as either true or false: (a) A hydrogen atom in the n = 3 state can emit light at only two specific wavelengths (b) a hydrogen atom in the n = 2 state is at a lower energy than one in the n = 1 state (c) the energy of an emitted photon equals the energy difference of the two states involved in the emission.

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Textbook Question

A stellar object is emitting radiation at 3.55 mm. a. What type of electromagnetic spectrum is this radiation? b. If a detector is capturing 3.2×108 photons per second at this wavelength, what is the total energy of the photons detected in 1.0 hour?

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Textbook Question

Molybdenum metal must absorb radiation with a minimum frequency of 1.09 * 1015 s - 1 before it can eject an electron from its surface via the photoelectric effect. (b) What wavelength of radiation will provide a photon of this energy?

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Textbook Question

Does the hydrogen atom 'expand' or 'contract' when an electron is excited from the n = 1 state to the n = 3 state?

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Textbook Question

A diode laser emits at a wavelength of 987 nm. (a) In what portion of the electromagnetic spectrum is this radiation found? (b) All of its output energy is absorbed in a detector that measures a total energy of 0.52 J over a period of 32 s. How many photons per second are being emitted by the laser?

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