Skip to main content
Ch.6 - Electronic Structure of Atoms
Chapter 6, Problem 29c

One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of 325 nm. (c) How many photons are in a 1.00 mJ burst of this radiation?

Verified Solution

Video duration:
3m
This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

Key Concepts

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

Photon Energy

The energy of a photon is determined by its wavelength and can be calculated using the equation E = hc/λ, where E is 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 in meters. For UV light at 325 nm, this relationship allows us to find the energy of a single photon.
Recommended video:
Guided course
01:40
Photon Energy Formulas

Conversion of Units

In chemistry, it is essential to convert units appropriately to ensure consistency in calculations. In this case, the energy burst of 1.00 mJ must be converted to joules (1 mJ = 0.001 J) to match the units used in the photon energy equation. This step is crucial for accurately determining the number of photons in the given energy burst.
Recommended video:
Guided course
01:56
Conversion Factors

Calculating Number of Photons

To find the number of photons in a given energy burst, divide the total energy by the energy of a single photon. This is expressed mathematically as N = E_total / E_photon, where N is the number of photons, E_total is the total energy in joules, and E_photon is the energy of one photon calculated from its wavelength. This calculation provides insight into the quantized nature of light.
Recommended video:
Guided course
01:22
Calculate Oxidation Numbers Example
Related Practice
Textbook Question

(b) What is the energy of one of these photons?

676
views
Textbook Question

(c) The laser pointer emits light because electrons in the material are excited (by a battery) from their ground state to an upper excited state. When the electrons return to the ground state, they lose the excess energy in the form of 532-nm photons. What is the energy gap between the ground state and excited state in the laser material?

2515
views
1
rank
Textbook Question

An AM radio station broadcasts at 1000 kHz and its FM partner broadcasts at 100 MHz. Calculate and compare the energy of the photons emitted by these two radio stations.

905
views
Textbook Question

One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of 325 nm. (d) These UV photons can break chemical bonds in your skin to cause sunburn—a form of radiation damage. If the 325-nm radiation provides exactly the energy to break an average chemical bond in the skin, estimate the average energy of these bonds in kJ>mol.

1890
views
Textbook Question

The energy from radiation can be used to rupture chemical bonds. A minimum energy of 192 kJ/mol is required to break the bromine–bromine bond in Br2. What is the longest wavelength of radiation that possesses the necessary energy to break the bond? What type of electromagnetic radiation is this?

1758
views
2
rank
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?

934
views