/*! 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 44 A lithium flame has a characteri... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A lithium flame has a characteristic red color due to emission of light of wavelength \(671 \mathrm{nm}\). What is the mass equivalence of \(1 \mathrm{~mol}\) of photons with this wavelength \(\left(1 \mathrm{~J}=1 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}^{2}\right) ?\)

Short Answer

Expert verified
The mass equivalence of 1 mole of photons with wavelength 671 nm is 1.98 x 10^-12 kg/mol.

Step by step solution

01

Determine the energy of one photon

Use the formula for energy of a photon, which is given by \[ E = \frac{hc}{\lambda} \] where \( E \) is the energy, \( h \) is Planck's constant \( 6.626 \times 10^{-34} \) J·s, \( c \) is the speed of light \( 3 \times 10^8 \) m/s, and \( \lambda \) is the wavelength in meters. First, convert the given wavelength from nanometers to meters: \( \lambda = 671 \) nm = \( 671 \times 10^{-9} \) m. Substitute the values into the equation: \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{671 \times 10^{-9}} \]
02

Calculate the energy of one photon

Perform the calculation:\[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{671 \times 10^{-9}} = 2.96 \times 10^{-19} \text{ J} \]
03

Determine the energy of 1 mole of photons

Use Avogadro’s number \( 6.022 \times 10^{23} \) to find the energy of 1 mole of photons. Multiply the energy per photon by Avogadro's number: \[ E_{\text{mole}} = 2.96 \times 10^{-19} \text{ J/photon} \times 6.022 \times 10^{23} \text{ photons/mol} \]
04

Calculate the energy of 1 mole of photons

Perform the multiplication:\[ E_{\text{mole}} = 2.96 \times 10^{-19} \times 6.022 \times 10^{23} = 1.78 \times 10^{5} \text{ J/mol} \]
05

Convert the energy to mass equivalence

Use Einstein's equation \( E = mc^2 \) to find the mass equivalence. Rearrange the equation to solve for mass \( m \): \[ m = \frac{E_{\text{mole}}}{c^2} \] Substitute \( E_{\text{mole}} = 1.78 \times 10^{5} \) J and \( c = 3 \times 10^8 \) m/s: \[ m = \frac{1.78 \times 10^{5}}{(3 \times 10^8)^2} \]
06

Calculate the mass equivalence

Perform the calculation: \[ m = \frac{1.78 \times 10^{5}}{9 \times 10^{16}} = 1.98 \times 10^{-12} \text{ kg} \text{/mol} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

photon energy calculation
Photons are tiny particles of light that carry energy. To determine the energy of a photon, we use the formula: \( E = \frac{hc}{\lambda} \). Here, \( E \) is the energy, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength.

For example, to find the energy of a photon with a wavelength of 671 nm, we first convert the wavelength from nanometers to meters. In this case, 671 nm = 671 x 10^{-9} meters.

Substituting in the values: \( E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{671 \times 10^{-9}} \), we perform the calculation to get the energy of one photon, which is \( 2.96 \times 10^{-19} \) joules (J).
Planck's constant
When discussing photon energy, Planck's constant \( (h) \) is a fundamental constant in physics that plays a key role. Its value is \( 6.626 \times 10^{-34} \) J·s.

Planck's constant is used to relate the energy of a photon to its frequency or wavelength. The formula \( E = h u \) (where \( u \) is frequency) or \( E = \frac{hc}{\lambda} \) shows this relationship.

Planck's constant is critical in quantum mechanics, as it signifies the smallest possible energy units (quanta) that can be emitted or absorbed as electromagnetic radiation.
mass-energy equivalence
Mass-energy equivalence is a concept in physics that shows how mass can be converted into energy and vice versa. This relationship is given by Einstein's famous equation: \( E = mc^2 \).

Here \( E \) is energy, \( m \) is mass, and \( c \) is the speed of light. This equation implies even a small amount of mass can be converted into a large amount of energy, because the speed of light \(( c \)) is a large number (approximately \( 3 \times 10^8 \) meters per second).

To find the mass equivalence of energy, we rearrange the equation to solve for mass: \( m = \frac{E}{c^2} \). Using this, if we know the energy, we can determine the corresponding mass.
Avogadro's number
Avogadro's number \( (N_A) \) is a fundamental constant in chemistry and physics that indicates the number of atoms, molecules, or particles in one mole of a substance. Its value is \( 6.022 \times 10^{23} \) particles per mole.

When dealing with photon energy, Avogadro's number helps us find the total energy in a mole of photons. For example, if one photon has an energy of \( 2.96 \times 10^{-19} \) J, the energy of one mole of these photons would be: \( 2.96 \times 10^{-19} \) J/photon \( \times 6.022 \times 10^{23} \) photons/mol = \( 1.78 \times 10^5 \) J/mol.

This is useful in fields like quantum chemistry and photochemistry, where reactions at the molecular level often involve calculations based on moles of photons.
Einstein's equation
Einstein's equation \( E = mc^2 \) is one of the most famous formulas in physics. It shows the relationship between energy \( (E) \) and mass \( (m) \) with a conversion factor \( c^2 \) (the speed of light squared).

This equation tells us that energy and mass are two forms of the same thing. Energy can be converted into mass and vice versa.

In practical terms, if we have the energy of a system or an object, we can find its equivalent mass. For example, the energy of one mole of photons ( \( 1.78 \times 10^5 \) J ) can be converted into mass: \( m = \frac{1.78 \times 10^5}{(3 \times 10^8)^2} = 1.98 \times 10^{-12} \) kg/mol. This shows the deep connection between mass and energy, and is foundational for understanding nuclear reactions and particle physics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Ozone formation in the upper atmosphere starts when oxygen molecules absorb UV radiation with wavelengths less than or equal to \(242 \mathrm{nm}\). Find the frequency and energy of the least energetic of these photons. (b) Ozone absorbs radiation with wavelengths in the range \(2200-2900\) A thus protecting organisms from this radiation. Find the frequency and energy of the most energetic of these photons.

Distinguish between an absorption spectrum and an emission spectrum. With which did Bohr work?

The following values are the only energy levels of a hypothetical one-electron atom: $$\begin{array}{ll} E_{6}=-2 \times 10^{-19} \mathrm{~J} & E_{5}=-7 \times 10^{-19} \mathrm{~J} \\\ E_{4}=-11 \times 10^{-19} \mathrm{~J} & E_{3}=-15 \times 10^{-19} \mathrm{~J} \\\ E_{2}=-17 \times 10^{-19} \mathrm{~J} & E_{1}=-20 \times 10^{-19} \mathrm{~J} \end{array}$$(a) If the electron were in the \(n=3\) level, what would be the highest frequency (and minimum wavelength) of radiation that could be emitted? (b) What is the ionization energy (in \(\mathrm{kJ} / \mathrm{mol}\) ) of the atom in its ground state? (c) If the electron were in the \(n=4\) level, what would be the shortest wavelength (in \(\mathrm{nm}\) ) of radiation that could be absorbed without causing ionization?

In the \(17^{\text {th }}\) century, Isaac Newton proposed that light was a stream of particles. The wave-particle debate continued for over 250 years until Planck and Einstein presented their ideas. Give two pieces of evidence for the wave model and two for the particle model.

Enormous numbers of microwave photons are needed to warm macroscopic samples of matter. A portion of soup containing \(252 \mathrm{~g}\) of water is heated in a microwave oven from \(20 .^{\circ} \mathrm{C}\) to \(98^{\circ} \mathrm{C}\), with radiation of wavelength \(1.55 \times 10^{-2} \mathrm{~m}\). How many photons are absorbed by the water in the soup?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.