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Chapter 8: Problem 4

The most intense line in the cerium spectrum is at 418.7 nm. (a) Determine the frequency of the radiation producing this line. (b) In what part of the electromagnetic spectrum does this line occur? (c) Is it visible to the eye? If so, what color is it? If not, is this line at higher or lower energy than visible light?

Short Answer

Expert verified
The frequency of the radiation producing the observed line in the cerium spectrum is approximately \(7.16 \times 10^{14} Hz\). It occurs in the ultraviolet (UV) part of the electromagnetic spectrum, and it is not visible to the human eye, as it has higher energy than visible light.

Step by step solution

01

Finding the frequency

The formula linking wavelength (\( \lambda \)) and frequency (\( v \)) of light is \( c = v \lambda \), where \( c \) is the speed of light (\(3.00 \times 10^8 m/s\)). The given wavelength has to be converted into meters by multiplying with \( 10^{-9} \). The frequency is then calculated by dividing the speed of light \( c \) by the converted wavelength (\( v = \frac{c}{\lambda} \))
02

Identify the spectrum

The electromagnetic spectrum ranges from radio waves to gamma rays. Given the wavelength for the cerium line, we can reference the wavelength ranges for each category of the electromagnetic spectrum to find that the cerium line falls within the ultraviolet (UV) category.
03

Determine the visibility and color

The visible part of the electromagnetic spectrum ranges from approximately 400nm to 700nm. While the cerium line is near this range, at 418.7nm, it technically falls within the ultraviolet (UV) category and therefore is not visible to the human eye. Its energy is higher than that of visible light, as shorter wavelengths correspond to higher energy.

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