From what I can gather the electrons are taken into account when calculating amu. Finally I have one more question, In the following question the author subtracts the mass of the alpha particle from the uranium nucleus, to get the mass of the thorium nucleus. Is the author just blatantly...
I see, so the nucleons lose some mass which is converted into energy and therefore the nucleus is held together albeit with less mass than the sum of its constituents. Does amu still take the electrons into account?
I am very confused with this question, firstly, I am unsure what is meant by the "unified atomic mass unit" I know that it is defined as "1/12 of the mass of an atom of carbon 12", but this sounds like it takes into account the electrons, i.e that is this means to me that unified atomic mass...
I understand the equation (path difference)/wavelength x 2pi = phase difference, but in this case I do not know how changing the distance of H from the source will affect the path difference. In addition to this, does minima refer to 0 amplitude (complete destructive interference) or does minima...
I am not really too sure where to start with this one, I would guess that the half life of the graph is very similar to the half life of the X to Y step, but at the same time I am unsure of how I would prove it. Any tips or reading material? Thanks!
Alright, so we use the word light to mean little to no mass (effectively 0 mass), and this means that the string doesn't need any force to compress. However it still needs force to extend right? And if we use the word light to describe a spring, as my textbook does, do the same things still apply?
I am happy with all the EPE being released, the issue was that I thought the particle needed some energy to squish the string down from a length of a to 0, but apparently you don't need any extra energy for that.