Final answer:
A photon with an energy of 4.903 x 10^(-20) J corresponds to ultraviolet radiation on the electromagnetic spectrum, determined by calculating its frequency using the photon energy equation with Planck's constant.
Step-by-step explanation:
If a photon is released from an atom with an energy of 4.903 x 10⁻²⁰ J, we first need to determine the frequency of the electromagnetic radiation using Planck's equation, which relates the energy (E) of a photon to its frequency (ν) through the relation E = hν, where h is Planck's constant (6.626 x 10⁻4 J·s).
First, we calculate the frequency of the photon:
E = hν
ν = E / h
ν = (4.903 x 10⁻²⁰ J) / (6.626 x 10⁻4 J·s)
ν = 7.39 x 10¹⁴ Hz
Next, using the electromagnetic spectrum, we know the ranges for different types of radiation by frequency:
- Infrared: < 10¹´ Hz
- Visible: 10¹´ Hz - 10¹µ Hz
- Ultraviolet: 10¹µ Hz - 10¹· Hz
- X-ray: > 10¹· Hz
Given the calculated frequency, the photon with an energy of 4.903 x 10⁻²⁰ J falls into the range of ultraviolet radiation. Therefore, the correct answer is b) Ultraviolet.