/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 Titanium metal requires a photon... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 34

Titanium metal requires a photon with a minimum energy of $6.94 \times 10^{-19} \mathrm{J}$ to emit electrons. (a) What is the minimum frequency of light necessary to emit electrons from titanium via the photoelectric effect? (b) What is the wavelength of this light? (c) Is it possible to eject electrons from titanium via the photoelectric effect? (b) What is the wavelength of this light? (c) Is it possible to eject electrons from titanium metal using visible light? (d) If titanium is irradiated with light of wavelength \(233 \mathrm{nm},\) what is the madimum possible kinetic energy of the emitted electrons?

Short Answer

Expert verified
The minimum frequency of light necessary to emit electrons from titanium is \(1.047 脳 10^{15} Hz\), and the corresponding wavelength is \(287 nm\). Since this is shorter than the range of visible light (\(400 nm - 700 nm\)), it's not possible to eject electrons from titanium using visible light. When titanium is irradiated with light of wavelength \(233 nm\), the maximum possible kinetic energy of the emitted electrons is \(1.60 脳 10^{-19} J\).

Step by step solution

01

1. Calculate the minimum frequency of light to emit electrons from titanium

To determine the minimum frequency of light necessary to emit electrons from titanium, we can use the Planck's equation, which states that the energy of a photon (E) is the product of Planck's constant (h) and the frequency of light (谓): \(E = h谓\) We're given the minimum necessary energy to emit electrons \(E = 6.94 脳 10^{-19} J\), and we know that Planck's constant \(h = 6.63 脳 10^{-34} Js\). Now solve for the frequency (谓). \(谓 = \frac{E}{h}\) \(谓 = \frac{6.94 脳 10^{-19} J}{6.63 脳 10^{-34} Js} = 1.047 脳 10^{15} Hz\) The minimum frequency of light necessary to emit electrons from titanium is \(1.047 脳 10^{15} Hz\).
02

2. Calculate the wavelength of the light

To find the wavelength of the light with the given frequency, we can use the speed of light equation: \(c = 位谓\) Where \(c = 3.00 脳 10^8 m/s\) is the speed of light in a vacuum, 位 is the wavelength, and 谓 is the frequency. We can solve for 位: \(位 = \frac{c}{谓}\) \(位 = \frac{3.00 脳 10^8 m/s}{1.047 脳 10^{15} Hz} = 2.87 脳 10^{-7} m\) The wavelength of this light is \(287 nm\).
03

3. Determine if visible light can eject electrons from titanium

Visible light has a wavelength range of approximately \(400 nm - 700 nm\). The wavelength of the light required to emit electrons from titanium is \(287 nm\), which is shorter than the minimum wavelength of visible light (400 nm). Therefore, it is not possible to eject electrons from titanium using visible light.
04

4. Calculate the maximum kinetic energy of emitted electrons

We are given that the wavelength of the light irradiating the titanium is \(233 nm\). First, we need to find the energy of the photons at this wavelength. Using the speed of light equation, \(c = 位谓\), we can solve for the frequency (谓): \(谓 = \frac{c}{位}\) \(谓 = \frac{3.00 脳 10^8 m/s}{233 脳 10^{-9} m} = 1.288 脳 10^{15} Hz\) Now we can use the Planck's equation to find the energy of these photons: \(E_\text{photon} = h谓\) \(E_\text{photon} = (6.63 脳 10^{-34} Js)(1.288 脳 10^{15} Hz) = 8.54 脳 10^{-19} J\) Since the energy of these photons is greater than the minimum required energy to emit electrons from titanium (\(6.94 脳 10^{-19} J\)), we can proceed to calculate the maximum possible kinetic energy of the emitted electrons. The excess energy from the photons will be converted into the kinetic energy of the electrons: \(K_\text{max} = E_\text{photon} - E_\text{min}\) \(K_\text{max} = 8.54 脳 10^{-19} J - 6.94 脳 10^{-19} J = 1.60 脳 10^{-19} J\) The maximum possible kinetic energy of the emitted electrons is \(1.60 脳 10^{-19} J\).

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Most popular questions from this chapter

(a) What experimental evidence is there for the electron having a "spin"? (b) Draw an energy-level diagram that shows the relative energetic positions of a \(1 s\) orbital and a \(2 s\) orbital. Put two electrons in the \(1 s\) orbital. (c) Draw an arrow showing the excitation of an electron from the \(1 s\) to the \(2 s\) orbital.

(a) What are the similarities of and differences between the \(1 \mathrm{~s}\) and \(2 s\) orbitals of the hydrogen atom? (b) In what sense does a \(2 p\) orbital have directional character? Compare the "directional" characteristics of the \(p_{x}\) and \(d_{x^{2}-y^{2}}\) orbitals. (That is, in what direction or region of space is the electron density concentrated?) (c) What can you say about the average distance from the nucleus of an electron in a \(2 s\) orbital as compared with a \(3 s\) orbital? (d) For the hydrogen atom, list the following orbitals in order of increasing energy (that is, most stable ones first): \(4 f, 6 s, 3 d, 1 s, 2 p\).

The discovery of hafnium, element number 72 , provided a controversial episode in chemistry. G. Urbain, a French chemist, claimed in 1911 to have isolated an element number 72 from a sample of rare earth (elements 58-71) compounds. However, Niels Bohr believed that hafnium was more likely to be found along with zirconium than with the rare earths. D. Coster and G. von Hevesy, working in Bohr's laboratory in Copenhagen, showed in 1922 that element 72 was present in a sample of Norwegian zircon, an ore of zirconium. (The name hafnium comes from the Latin name for Copenhagen, Hafnia). (a) How would you use electron configuration arguments to justify Bohr's prediction? (b) Zirconium, hafnium's neighbor in group \(4 \mathrm{~B}\), can be produced as a metal by reduction of solid \(\mathrm{ZrCl}_{4}\) with molten sodium metal. Write a balanced chemical equation for the reaction. Is this an oxidation-reduction reaction? If yes, what is reduced and what is oxidized? (c) Solid zirconium dioxide, \(\mathrm{ZrO}_{2}\), reacts with chlorine gas in the presence of carbon. The products of the reaction are \(\mathrm{ZrCl}_{4}\) and two gases, \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) in the ratio \(1: 2\). Write a balanced chemical equation for the reaction. Starting with a 55.4-g sample of \(\mathrm{ZrO}_{2}\), calculate the mass of \(\mathrm{ZrCl}_{4}\) formed, assuming that \(\mathrm{ZrO}_{2}\) is the limiting reagent and assuming \(100 \%\) yield. (d) Using their electron configurations, account for the fact that \(\mathrm{Zr}\) and \(\mathrm{Hf}\) form chlorides \(\mathrm{MCl}_{4}\) and oxides \(\mathrm{MO}_{2}\).

(a) What is the relationship between the wavelength and the frequency of radiant energy? (b) Ozone in the upper atmosphere absorbs energy in the \(210-230-\mathrm{nm}\) range of the spectrum. In what region of the electromagnetic spectrum does this radiation occur?

The rays of the Sun that cause tanning and burning are in the ultraviolet portion of the electromagnetic spectrum. These rays are categorized by wavelength. So-called UV-A radiation has wavelengths in the range of \(320-380 \mathrm{~nm}\), whereas UV-B radiation has wavelengths in the range of 290-320 \(\mathrm{nm}\). (a) Calculate the frequency of light that has a wavelength of \(320 \mathrm{~nm}\). (b) Calculate the energy of a mole of \(320-\mathrm{nm}\) photons. (c) Which are more energetic, photons of UV-A radiation or photons of UV-B radiation? (d) The UV-B radiation from the Sun is considered a greater cause of sunburn in humans than is UV-A radiation. Is this observation consistent with your answer to part (c)?
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