Answer:
because the water in a bottle has been purified
Explanation:
In order to answer this question we might first want to think about what is electromagnetic radiation. In essence it’s light, just some of the wavelengths are too long or too short for us to see.
We can think about it as two oscillating sinusoidal (goes up and down) waves, one is electric, the other is magnetic.
Because we’re dealing in waves, that means we can calculate their frequency, wavelength, amplitude (brightness) and period.
To calculate it we can use E=hc/lambda
Where E = jewels of energy
h = Planck’s constant
c = speed of light
Lambda = wavelength
It doesn’t really matter for this question what those things mean, just note that it takes more energy to have a shorter wavelength, or less energy to have a longer wavelength.
So now we can answer the question. Light of a longer wavelength has less energy than that of a shorter wavelength. So, when long wavelengths are absorbed by matter (atoms) they will give those atoms less energy. So, either it will pass through the object entirely or it will make the atoms vibrate a little bit more than they already are and we call that thermal energy, or heat.
If high energy wavelengths are passing through matter then they will be giving those atoms a lot of energy, sometimes even ionizing the atoms.
Which, if you’re a living thing can be very bad for your cells.
I hope that helps.
Acid-base tiltrations has DNA that can help in the process of neutralization. not so sure about it
. The energy of shells in a hydrogen atom is calculated by the formula E = -Eo/n^2 where n is any integer, and Eo = 2.179X10^-18 J. So, the energy of a ground state electron in hydrogen is:
E = -2.179X10^-18 J / 1^2 = -2.179X10^-21 kJ
Consequently, to ionize this electron would require the input of 2.179X10^-21 kJ
2. The wavelength of a photon with this energy would be:
Energy = hc/wavelength
wavelength = hc/energy
wavelength = 6.626X10^-34 Js (2.998X10^8 m/s) / 2.179X10^-18 J = 9.116X10^-8 m
Converting to nanometers gives: 91.16 nm
3. Repeat the calculation in 1, but using n=5.
4. Repeat the calculation in 2 using the energy calculated in 3.
