Why Are My Particle Physics Calculations Incorrect?

AI Thread Summary
The discussion centers on a student's difficulty in obtaining correct answers for particle physics calculations, specifically when calculating the specific charge of a carbon ion. The student correctly identifies the formula but initially fails to use parentheses in their calculator, leading to an incorrect result. After receiving advice to use brackets, the student successfully calculates the correct answer. Additionally, there is a query about how to set the calculator to display results in scientific notation, with suggestions provided based on different calculator models. Proper use of parentheses and understanding calculator settings are emphasized as key to resolving calculation issues.
bobo343
Messages
4
Reaction score
0
I'm currently studying physics at A-level, and for particle physics I'm fine about the theory of the equations, I remember what equations to use etc, but I can't seem to get the right answers from my calculator.

For example, when working out the specific charge of a carbon ion with Z = 6, A = 13 and a charge of 2+, the necessary equation would be Specific Charge = charge/mass.

Which is: 2 x 1.60 x 10^-19 / 6 x 1.67 x 10^-27 + 7 x 1.67 x 10^-27 + 4 x 9.11 x 10^-31

When shortened is: 2 x 1.60 x 10^-19 / 13 x 1.67 x 10^-27

Which finally should = 1.47 x10^7 C kg^-1

But when I enter 2 x 1.60 x 10^-19 / 13 x 1.67 x 10^-27 into my calculator, I get a vastly different answer of 4.110769231 x 10^-47

Is there something I'm doing wrong? We haven't done particle physics in a while and I might be missing a vital step.

Any help will be greatly appreciated.
 
Physics news on Phys.org
I suggest you use brackets around the terms to ensure your calculator is doing exactly what you want it to.
(2x1.6x10^-19) / (13x1.67x10^-27)
and try again now.
 
Alright, with the brackets I get 14739751.27.

Is there any way I can set my calculator to give me the answer with the x10^n? (n just being whatever power)

Thanks for the help.
 
It depends on the calculator, but mine has a "scientific" setting, where the results in exponentials come up as e.g. 3.4567Exp8.
It's probably not a bad idea to practise putting things in the right scientific notation anyway though.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top