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Ch.6 - Electronic Structure of Atoms
Chapter 6, Problem 34

Titanium metal requires light with a maximum wavelength of 286 nm to emit electrons. (a) What is the minimum energy of the photons necessary to emit electrons from titanium via the photoelectric effect? (b) If titanium is irradiated with light of wavelength 276 nm, what is the maximum possible kinetic energy of the emitted electrons?

Verified step by step guidance
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Step 1: Understand the photoelectric effect, which states that electrons are emitted from a material when it is exposed to light of a certain minimum frequency or maximum wavelength. The energy of the photons must be greater than or equal to the work function of the material.
Step 2: Use the equation for the energy of a photon, E = \frac{hc}{\lambda}, where h is Planck's constant (6.626 \times 10^{-34} \text{ J s}), c is the speed of light (3.00 \times 10^8 \text{ m/s}), and \lambda is the wavelength of the light. Calculate the minimum energy of the photons using the maximum wavelength of 286 nm.
Step 3: Convert the wavelength from nanometers to meters for calculation purposes. Remember that 1 nm = 1 \times 10^{-9} m.
Step 4: For part (b), calculate the energy of the photons with a wavelength of 276 nm using the same formula, E = \frac{hc}{\lambda}.
Step 5: Determine the maximum kinetic energy of the emitted electrons by using the equation KE_{max} = E_{photon} - \phi, where \phi is the work function of titanium, which is the energy calculated in Step 2.

Key Concepts

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

Photoelectric Effect

The photoelectric effect is a phenomenon where electrons are emitted from a material when it absorbs light of sufficient energy. This effect demonstrates the particle nature of light, where photons must have a minimum energy, determined by the material's work function, to dislodge electrons. The relationship between the energy of the photons and the emitted electrons is crucial for understanding how light interacts with matter.
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Photon Energy Calculation

The energy of a photon can be calculated using the equation E = hc/λ, where E is the energy, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the light in meters. This relationship shows that shorter wavelengths correspond to higher energy photons, which is essential for determining whether the light can cause electron emission in the photoelectric effect.
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Kinetic Energy of Emitted Electrons

The maximum kinetic energy (KE) of emitted electrons in the photoelectric effect can be calculated using the equation KE = E - φ, where E is the energy of the incoming photon and φ is the work function of the material. This concept is important for understanding how excess energy from the photon translates into the kinetic energy of the emitted electrons, allowing for the calculation of their maximum speed and behavior.
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Related Practice
Textbook Question

Molybdenum metal must absorb radiation with an energy higher than 7.22 * 10-19 J ('energy threshold') before it can eject an electron from its surface via the photoelectric effect. (a) What is the frequency threshold for emission of electrons?

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Textbook Question

Molybdenum metal must absorb radiation with an energy higher than 7.22 * 10-19 J ('energy threshold') 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

Molybdenum metal must absorb radiation with an energy higher than 7.22 * 10-19 J ('energy threshold') before it can eject an electron from its surface via the photoelectric effect. (c) If molybdenum is irradiated with light of wavelength of 240 nm, what is the maximum possible velocity of the emitted electrons?

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

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.

Textbook Question

Is energy emitted or absorbed when the following electronic transitions occur in hydrogen? (a) from n = 3 to n = 2 (c) an electron adds to the H+ ion and ends up in the n = 2 shell?