00:01
In this problem, we are given a line spectrum of mercury, with lines at 253 .652 nanometers, 365 .05915 nanometers, 404 .833 nanometers, and 1 ,13 .975 nanometers.
00:18
For part a, we are asked to find which one of the line represents the most energetic light.
00:23
So we know that energy is inversely proportional to wavelength.
00:26
This means that as wavelength decreases, energy.
00:31
Increases.
00:33
By this logic, the line with the smallest wavelength would have the highest energy.
00:38
Therefore, the line at 253 .652 nanometers since it has the smallest wavelength would have the highest energy.
01:00
For per b we are asked to calculate the frequency of the most prominent line, which is 253 .652 nanometers.
01:08
So we know that the speed of light equals frequency times wavelength.
01:16
So we want to isolate frequency.
01:20
So you would do this by dividing both sides by wavelength...