sorry to be a bit daft, but still cat really get to the end.
Let me summarize wat i understand so far first.
The maximum possible uncertainty in E of a particle can be m_{0}c^{2}, because anything beyond that it would have enough energy to form another particle.
and from...
ah ok thanks tahnks, that's a bit helpful.
But i am still a bit unclear. That mean i don't have to do any rigourous maths like diffrenciations?
and also i am not clear on the part they said, "when \Delta p exceeds mc, the uncertainty in energy is greater than mc^{2}" Why?
Homework Statement
Show that the smallest possible uncertainty in the position of an electron whose speed is given by \beta=v/c is
\Delta x_{min}=\frac{h}{4\pi m_{0}c}(1-\beta^{2})^{1/2}
Homework Equations
\Delta x \Delta p=\frac{h}{4\pi}
p=mv= \frac{m_{0}}{\sqrt{1-\beta^{2}}}v...
Actually according to the prof, the grating was mistakenly bought this experiment. I suppose the original plan was to get a grating that would not require calibrating cos the lab manual never said anything about calibrating the grating. But since we end up with this, prof decides to turn this...
Hi, i am new here, so am not really sure wheather i am asking this qus at the right place. I have a Lab based qus. I just did the atomic spetra analysis using a diffraction grating, analysing hydrogen gas.
But but when i calculate the wavelengths of the lines, its way off from the literature...